1. Compound Angle Formula

Hi everyone, hope you are all ok this evening.

Please could someone help with the working out of the following questions...

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1. Using compound angle formula show that

Sin(x+90 º) + Cos(x-180 º) = 0

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2. Using compound angle formula show that

Sin(A-B) + Sin(A+B) = 2Sin A Cos B

Regards
Paul

2. Originally Posted by duckegg911
Hi everyone, hope you are all ok this evening.

Please could someone help with the working out of the following questions...

________________________________________
1. Using compound angle formula show that

Sin(x+90 º) + Cos(x-180 º) = 0

________________________________________
2. Using compound angle formula show that

Sin(A-B) + Sin(A+B) = 2Sin A Cos B

Regards
Paul

im guessing when you say "compound" you mean Double Angle forumla? I guess in Canada they call it something different.
You have to solve the right side of the equation. 2(Sina+cosb) = (sinacosb)+(sinacosb)
then you have to simplify using the regular formulas i think

I took this a few months ago it is something like this, im sure someone else can help more. Its a start though

3. Hello, Paul!

You are expected to know the Compound Angle Formulas:

. . $\displaystyle \sin(A \pm B) \;=\;\sin A\cos B \pm \sin B\cos A$

. . $\displaystyle \cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B$

$\displaystyle (1)\;\;\sin(x+90^o)+\cos(x-180^o) \:=\: 0$
$\displaystyle \sin(x+90^o) + \cos(x-180^o)$

. . $\displaystyle = \;\bigg[\sin x\cos90^o + \cos x\sin90^o\bigg] + \bigg[\cos x\cos180^o + \sin x\sin180^o\bigg]$

. . $\displaystyle = \;\bigg[\sin x\cdot 0 + \cos x\cdot 1\bigg] + \bigg[\cos x(-1) + \sin x\cdot 0\bigg]$

. . $\displaystyle = \;0 + \cos x - \cos x + 0$

. . $\displaystyle =\;\;0$

$\displaystyle (2)\;\;\sin(A-B) + \sin(A+B) \:=\:2\sin A\cos B$
We have: .$\displaystyle \begin{array}{cccc}\sin(A-B) &=& \sin A\cos B - \sin B\cos A & {\color{blue}[1]} \\ \sin(A+B) &=& \sin A\cos B + \sin B\cos A & {\color{blue}[2]} \end{array}$

Add [1] and [2]: . $\displaystyle \sin(A-B) + \sin(A+B) \;=\;2\sin A\cos B$

4. Thanks very much for question 1...

I fully understand how you did that now!

In question 2 can you show me how you added the two equations together?

Best Regards

+ Many Thanks
Paul

5. Can anyone expand on the soloution for example number two please?

Regards
Paul

6. Please may I have assistance with this problem

Using compound angle formula show that

cos 2A = 1 - 2sin²A

Many thanks
Dan

7. (A+B) =cosAcosB - sinAsinB
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Here A=B

cos(2A) =cos^2A - sin^2A
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Do you remember that

cos^2A +sin^2A =1

this will give

cos^2A = 1- sin^2A

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Hence

$\displaystyle cosA = cos^2A -sin^2A = 1- sin^2A -sin^2A = 1-2sin^2A$
__________________________________________________ ____________

$\displaystyle sin(A+B) +sin(A-B) = (sinAcosB +cosAsinB )+ (sinAcosB-cosAsinB)$

$\displaystyle sinAcosB +sinAcossB {\color{red}+cosAsinB-cosAsinB}= 2sinAcosB$

Things in red get cancelled