reduction of a cosΘ+/_ b sinΘ

**jasonlewiz** · September 21st 2009, 07:10 AM

reduct this problem..

1/2 sinθ +

/2 cos Θ

to the form

a. r sin (Θ+Φ)
b. r cos(Θ+Φ)

**skeeter** · September 21st 2009, 07:15 AM

Originally Posted by jasonlewiz

reduct this problem..

1/2 sinθ +

/2 cos Θ

to the form

a. r sin (Θ+Φ)
b. r cos(Θ+Φ)

$<br /> \cos\left(\frac{\pi}{3}\right) \sin(\theta) + \sin\left(\frac{\pi}{3}\right) \cos(\theta)<br />$

or

$<br /> \sin\left(\frac{\pi}{6}\right) \sin(\theta) + \cos\left(\frac{\pi}{6}\right) \cos(\theta)<br />$

finish up by using the sum/difference identities for sine and cosine

**Grandad** · September 21st 2009, 09:50 AM

Hello jasonlewiz

Originally Posted by jasonlewiz

reduct this problem..

1/2 sinθ +

/2 cos Θ

to the form

a. r sin (Θ+Φ)
b. r cos(Θ+Φ)

skeeter's solution is all you need for this particular problem. But, in case you don't spot that $\tfrac12$ can be written as $\cos\tfrac{\pi}{3}$ and that $\tfrac{\sqrt3}{2}$ is the sine of the same angle ( $\tfrac{\pi}{3}$ ), here's the more general method for handling this type of expression:

$\tfrac12\sin\theta + \tfrac{\sqrt3}{2}\cos\theta = r \sin(\theta + \phi)$

$= r\sin\theta\cos\phi + r\cos\theta\sin\phi$

Compare coefficients of $\sin\theta$ and $\cos\theta$ :

$\tfrac12 = r \cos\phi$ (1)

$\tfrac{\sqrt3}{2} = r\sin\phi$ (2)

Divide (2) by (1): $\tan\phi = \sqrt3$

$\Rightarrow \phi = \tfrac{\pi}{3}$

Square (1) and (2) and add:

$\tfrac14 + \tfrac34 = r^2(\sin^2\phi + \cos^2\phi)=r^2$

$\Rightarrow r = 1$ (taking the positive root)

So there are the values of $r$ and $\phi$ .

You can do part (b) in exactly the same way, but starting with $\tfrac12\sin\theta + \tfrac{\sqrt3}{2}\cos\theta = r \cos(\theta + \phi)$ . (In fact, it's slightly easier to start with $\tfrac12\sin\theta + \tfrac{\sqrt3}{2}\cos\theta = r \cos(\theta - \phi)$ since $\phi$ then turns out to be a positive angle.)

Grandad

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reduction of a cosΘ+/_ b sinΘ

Handling expressions of the form a sin x + b cos x

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