Show that...

[hcinic]

Having trouble with this:

If cos(a+B)=0, show that sin(a+2B)=sina

I know my addition formulas, but I also know that I am missing something because when I expand them I get stuck there. There must be a rule or identity that I'm not seeing. Any advice or help?

 

[hcinic]

cosacosB-sinasinB=0

show that

sinacos2B+cosasin2B=sina

I'm thinking expand cos2B and sin2B

 

[Deleted member 4993]

cosacosB-sinasinB=0

show that

sinacos2B+cosasin2B=sina

I'm thinking expand cos2B and sin2B

Yes.. go ahead expand cos2B = 2cos²B - 1

 

[hcinic]

I did, expanded cos2B all three ways, including what you suggested and seems to not get me where I think I need to be. Also expanded sin2B=2sinBcosB. Still working on it though, not giving up by any means. I'll get it :wink:

 

[hcinic]

I keep thinking I should be seeing something in cos(a+B)=0 that will help me show that sin(a+2B)=sina

 

[Deleted member 4993]

Okay... change of plan - however, I am not getting the answer either...

cos(A+B) = 0 → sin(A+B) = ±1

sin(A+B+B) = sin(A+B)*cosB + cos(A+B)*sinB = ± cosB

Are you sure you posted the problem correctly?

 

[hcinic]

Yes, it is posted correctly.

Given cos(alpha + beta)=0

show that sin(alpha + 2beta)=sin alpha

 

[hcinic]

Okay... change of plan - however, I am not getting the answer either...

cos(A+B) = 0 → sin(A+B) = ±1

sin(A+B+B) = sin(A+B)*cosB + cos(A+B)*sinB = ± cosB

Are you sure you posted the problem correctly?

I think you are right, because doesn't cosB=sina?

 

[Deleted member 4993]

\(\displaystyle Let\ k\ be\ an\ integer.\)

\(\displaystyle cos( \alpha + \beta ) = 0 \implies \alpha + \beta = (2k - 1) \left( \dfrac{ \pi }{2} \right) \implies \beta = (2k - 1) \left( \dfrac{ \pi }{2} \right) - \alpha \implies\)

\(\displaystyle \alpha + 2 \beta = \alpha + 2\left[(2k - 1) \left( \dfrac{ \pi }{2} \right) - \alpha \right] = (2k - 1) \pi - \alpha \implies\)

\(\displaystyle sin( \alpha + 2 \beta ) = sin[ (2k - 1) \pi - \alpha ] = sin[ (2k - 1) \pi] * cos( \alpha) - cos[ (2k - 1) \pi ] * sin( \alpha) = 0 * cos( \alpha ) - (-1) * sin( \alpha) = sin( \alpha ).\)

Check my work. I have not done trig in decades.

This is the solution .....

 

[hcinic]

Nice. Thanks JeffM! This also helps with where Khan was headed, or a different route, which I believe will also work:

cofunction identities - cos(pi/2-theta)=sin theta and sin(pi/2-theta)=cos theta