Question

Help,
Use co-functions identities to find the solution of cos θ = sin 5θ. Assume θ is acute.

Answer 1

Recall that sine and cosine are cofunctions.
So the cosine of an angle will be equivalent to the sine of the compliment of that angle.\[\Large\rm \cos (x)=\sin\left(90-x\right)\]Or in radians,\[\Large\rm \cos (x)=\sin\left(\frac{\pi}{2}-x\right)\]

Answer 2

For our problem we can say that,\[\Large\rm \sin(5\theta)\]is the same as,\[\Large\rm \cos\left(\frac{\pi}{2}-5\theta\right)\]

Answer 3

So our problem becomes,\[\Large\rm \cos \theta=\cos\left(\frac{\pi}{2}-5\theta\right)\]

Answer 4

So I guess we can say that these two sides are equivalent when the angles are equivalent (allowing for rotation).\[\Large\rm \implies\qquad \theta=\frac{\pi}{2}-5\theta+2k \pi,\qquad\qquad k=\pm(0,1,2,...)\]Combine like-terms,\[\Large\rm 6\theta=\frac{\pi}{2}+2k \pi\]Divide by 6,\[\Large\rm \theta=\frac{\pi}{12}+k\frac{4\pi}{12}\]

Answer 5

And I guess if we're assuming theta is `acute`, then we can only use k=0 and k=1. Those will give us small enough angles.

pi/12 and 5pi/12 or something like that, ya?

Answer 6

Hmmm there is a pi/8 solution also though :d
I'm trying to figure out where that is coming from...
I missed something >.<

Answer 7

it says the answer is (theta) = 15 degrees

Answer 8

pi/12 = 15 degrees.
5pi/12 = 75 degrees. <-- I don't see why they wouldn't include* this as a solution D: it's still acute.