Question

Find sin2x, cos2x, and tan2x if cosx=3/5 and x terminates in quadrant IV.

Answer 1

\[\cos x = \frac{3}{5} = \frac{adj}{hyp}\Rightarrow adj = 3, hyp = 5 \\
opp = \sqrt{hyp^2 - adj^2} = \sqrt{25 - 9} = 4 \textrm{ (also it's a 3-4-5 triangle)} \\
\sin x = \frac{opp}{hyp} = \frac{4}{5} \\
\sin 2x = 2\sin x\cos x = 2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right) = \frac{24}{25} \\
\]
You should be able to solve the pthers in a similar fashion.

Answer 2

Thanks. But how does Sin2x become sinx cosx? I don't understand that part.

Answer 3

Double angle formula:
\[\sin 2x = 2 \sin x \cos x\]

Answer 4

If you remember the formula for the sin of the sum of two angles:
\[\sin(A+B) = \sin A \cos B + \cos A \sin B\]and let \[A = x, B=x\]\[\sin 2x = \sin(x+x) = \sin x \cos x + \cos x \sin x \]\[= \sin x \cos x + \sin x\cos x =2 \sin x\cos x\]

Answer 5

Thanks for the help. This part of precalc is killin me right now.

Answer 6

It's worth taking the time to really learn the identities...you'll make extensive use of this stuff in calculus.