# [SOLVED] Verifying Trig identities

• Feb 5th 2009, 04:22 PM
jamespk07
[SOLVED] Verifying Trig identities
Umm the question is

sin(3pi-x) = sinx

well i figured to use the sum and difference formulas but after i change it to the formula i get stuck... and i dont know where elso to go

same with these other ones-

cos(5pi/4-x) = square root of -2/2(cos x + sin x)
• Feb 5th 2009, 05:16 PM
Soroban
Hello, jamespk07!

You're expected to know some standard trig values . . .

Quote:

Prove: .$\displaystyle \sin(3\pi-x) \:= \:\sin x$

$\displaystyle \sin(3\pi-x) \:=\:\sin(3\pi)\cos(x) - \sin(x)\cos(3\pi)$

. . . . . . . $\displaystyle = \;0\cdot\cos(x) - \sin(x) \cdot(-1)$

. . . . . . . $\displaystyle = \;\sin x$

Quote:

$\displaystyle \cos\left(\tfrac{5\pi}{4}-x\right) \:=\:\text{-}\tfrac{\sqrt{2}}{2}(\cos x + \sin x)$

$\displaystyle \cos\left(\tfrac{5\pi}{4} - x\right) \;=\;\cos\left(\tfrac{5\pi}{4}\right)\cos x - \sin\left(\tfrac{5\pi}{4}\right)\sin x$

. . . . . . . . $\displaystyle = \;\left(\text{-}\tfrac{\sqrt{2}}{2}\right)\cos x + \left(\text{-}\tfrac{\sqrt{2}}{2}\right)\sin x$

. . . . . . . . $\displaystyle = \;\text{-}\tfrac{\sqrt{2}}{2}(\cos x + \sin x)$