Question

write sin x + sin 5x as a product

Answer 1

Hint: Use the identity


\[\large \sin(u) + \sin(v) = 2\sin\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right)\]

Answer 2

thanks!

Answer 3

you're welcome, tell me what you get

Answer 4

idk if i'm doing this right! sin (1) + sin (5) =2sin3cos-3

Answer 5

\[\large \sin(u) + \sin(v) = 2\sin\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right)\]

\[\large \sin(x) + \sin(5x) = 2\sin\left(\frac{x+5x}{2}\right)\cos\left(\frac{x - 5x}{2}\right)\]

\[\large \sin(x) + \sin(5x) = 2\sin\left(\frac{6x}{2}\right)\cos\left(\frac{-4x}{2}\right)\]

\[\large \sin(x) + \sin(5x) = 2\sin\left(3x\right)\cos\left(-2x\right)\]

\[\large \sin(x) + \sin(5x) = 2\sin\left(3x\right)\cos\left(2x\right)\]

Answer 6

btw I found that identity here

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 7

thanksssss

Answer 8

sure thing

Answer 9

Now that i'm looking at it, it wasn't difficult at all! The only difficult part was knowing what formula to use.

Answer 10

yeah it's not too bad

Answer 11

would you be willing to help me solve csc (cos^-1 (15/17))

Answer 12

draw a right triangle and mark one of the angles (that's not a right angle) as \(\large \theta\)

Answer 13

now cos^-1(15/17) refers to the fact that there is some angle \(\large \theta\) such that

\(\large \cos(\theta) = \frac{15}{17}\)

Answer 14

since cos = adj/hyp, we know that

Answer 15

how would you find the opposite side