Proving $1-cos2a+2sin a sin3a = 2 sin^22a$ and $sin^22a - sin^2a = sin3asin a$
Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.
1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$
I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
$$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
but I am lost after this point.
2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$
I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.
That brings me to:
$$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
or
$$sin^2a(4 cos^2a -1) = sin3asin a$$
but I cannot find the solution.
So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?
Thanks for your attention. I’m looking forward to your reply.
trigonometry
edited Nov 28 at 22:38
Blue
47.5k870150
asked Nov 28 at 21:33
Toa Narumi
91
add a comment |
Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.
1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$
I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
$$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
but I am lost after this point.
2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$
I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.
That brings me to:
$$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
or
$$sin^2a(4 cos^2a -1) = sin3asin a$$
but I cannot find the solution.
So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?
Thanks for your attention. I’m looking forward to your reply.
trigonometry
edited Nov 28 at 22:38
Blue
47.5k870150
asked Nov 28 at 21:33
Toa Narumi
91
add a comment |
Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.
1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$
I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
$$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
but I am lost after this point.
2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$
I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.
That brings me to:
$$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
or
$$sin^2a(4 cos^2a -1) = sin3asin a$$
but I cannot find the solution.
So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?
Thanks for your attention. I’m looking forward to your reply.
trigonometry
edited Nov 28 at 22:38
Blue
47.5k870150
asked Nov 28 at 21:33
Toa Narumi
91
Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.
1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$
I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
$$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
but I am lost after this point.
2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$
I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.
That brings me to:
$$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
or
$$sin^2a(4 cos^2a -1) = sin3asin a$$
but I cannot find the solution.
So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?
Thanks for your attention. I’m looking forward to your reply.
trigonometry
trigonometry
edited Nov 28 at 22:38
Blue
47.5k870150
asked Nov 28 at 21:33
Toa Narumi
91
edited Nov 28 at 22:38
Blue
47.5k870150
asked Nov 28 at 21:33
Toa Narumi
91
edited Nov 28 at 22:38
Blue
47.5k870150
edited Nov 28 at 22:38
Blue
47.5k870150
edited Nov 28 at 22:38
Blue
47.5k870150
47.5k870150
asked Nov 28 at 21:33
Toa Narumi
91
asked Nov 28 at 21:33
Toa Narumi
91
asked Nov 28 at 21:33
Toa Narumi
91
91
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Hint:
Linearise the r.h.s.:
$$2sin^22a=1-cos 4a,$$
and, in the l.h.s., $;2sin asin 3a$.
answered Nov 28 at 21:47
Bernard
117k638111
add a comment |
- Hint:
Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
$$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
$$sin(3alpha)=3sinalpha-4sin^3alpha$$
SOLUTION
$$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$
- SOLUTION
$$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$
answered Nov 28 at 22:06
Dr. Mathva
929315
add a comment |
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin asin3a=cos(3-1)a-cos(3+1)a$$
Then use $cos2y=1-2sin^2y$
For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $
See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
add a comment |
HINT
For the first one use that
$$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$
then
$$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$
$$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
$$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$
and recall that $cos (2x)=1-2sin^2 x$.
Can you proceed also for the second one, using the same identities?
answered Nov 28 at 21:41
gimusi
1
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
Linearise the r.h.s.:
$$2sin^22a=1-cos 4a,$$
and, in the l.h.s., $;2sin asin 3a$.
answered Nov 28 at 21:47
Bernard
117k638111
add a comment |
Hint:
Linearise the r.h.s.:
$$2sin^22a=1-cos 4a,$$
and, in the l.h.s., $;2sin asin 3a$.
answered Nov 28 at 21:47
Bernard
117k638111
add a comment |
Hint:
Linearise the r.h.s.:
$$2sin^22a=1-cos 4a,$$
and, in the l.h.s., $;2sin asin 3a$.
answered Nov 28 at 21:47
Bernard
117k638111
Hint:
Linearise the r.h.s.:
$$2sin^22a=1-cos 4a,$$
and, in the l.h.s., $;2sin asin 3a$.
answered Nov 28 at 21:47
Bernard
117k638111
answered Nov 28 at 21:47
Bernard
117k638111
answered Nov 28 at 21:47
Bernard
117k638111
answered Nov 28 at 21:47
Bernard
117k638111
117k638111
add a comment |
add a comment |
- Hint:
Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
$$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
$$sin(3alpha)=3sinalpha-4sin^3alpha$$
SOLUTION
$$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$
- SOLUTION
$$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$
answered Nov 28 at 22:06
Dr. Mathva
929315
add a comment |
- Hint:
Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
$$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
$$sin(3alpha)=3sinalpha-4sin^3alpha$$
SOLUTION
$$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$
- SOLUTION
$$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$
answered Nov 28 at 22:06
Dr. Mathva
929315
add a comment |
- Hint:
Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
$$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
$$sin(3alpha)=3sinalpha-4sin^3alpha$$
SOLUTION
$$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$
- SOLUTION
$$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$
answered Nov 28 at 22:06
Dr. Mathva
929315
- Hint:
Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
$$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
$$sin(3alpha)=3sinalpha-4sin^3alpha$$
SOLUTION
$$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$
- SOLUTION
$$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$
answered Nov 28 at 22:06
Dr. Mathva
929315
answered Nov 28 at 22:06
Dr. Mathva
929315
answered Nov 28 at 22:06
Dr. Mathva
929315
answered Nov 28 at 22:06
Dr. Mathva
929315
929315
add a comment |
add a comment |
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin asin3a=cos(3-1)a-cos(3+1)a$$
Then use $cos2y=1-2sin^2y$
For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $
See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
add a comment |
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin asin3a=cos(3-1)a-cos(3+1)a$$
Then use $cos2y=1-2sin^2y$
For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $
See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
add a comment |
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin asin3a=cos(3-1)a-cos(3+1)a$$
Then use $cos2y=1-2sin^2y$
For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $
See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin asin3a=cos(3-1)a-cos(3+1)a$$
Then use $cos2y=1-2sin^2y$
For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $
See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
answered Nov 29 at 4:58
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
HINT
For the first one use that
$$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$
then
$$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$
$$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
$$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$
and recall that $cos (2x)=1-2sin^2 x$.
Can you proceed also for the second one, using the same identities?
answered Nov 28 at 21:41
gimusi
1
add a comment |
HINT
For the first one use that
$$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$
then
$$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$
$$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
$$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$
and recall that $cos (2x)=1-2sin^2 x$.
Can you proceed also for the second one, using the same identities?
answered Nov 28 at 21:41
gimusi
1
add a comment |
HINT
For the first one use that
$$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$
then
$$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$
$$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
$$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$
and recall that $cos (2x)=1-2sin^2 x$.
Can you proceed also for the second one, using the same identities?
answered Nov 28 at 21:41
gimusi
1
HINT
For the first one use that
$$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$
then
$$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$
$$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
$$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$
and recall that $cos (2x)=1-2sin^2 x$.
Can you proceed also for the second one, using the same identities?
answered Nov 28 at 21:41
gimusi
1
edited Nov 28 at 21:59
answered Nov 28 at 21:41
gimusi
1
answered Nov 28 at 21:41
gimusi
1
answered Nov 28 at 21:41
gimusi
1
1
add a comment |
add a comment |
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