# identity 2

• Jul 27th 2009, 10:47 PM
william
identity 2
1. prove that cos $\displaystyle (\frac{3\pi}{2}+x)= sin x$

2. is $\displaystyle sin 2x sin x- cos x = \frac{\sqrt3}{4}$ is true for $\displaystyle \frac{7\pi}{6}$? Explain
• Jul 27th 2009, 10:51 PM
pomp
Quote:

Originally Posted by william
1. prove that cos $\displaystyle (\frac{3\pi}{2}+x)= sin x$

$\displaystyle \cos{\left(A+B\right)} = \cos{A}\cos{B} - \sin{A}\sin{B}$

So substitute what you have in question 1 into this formula.
• Jul 27th 2009, 11:26 PM
red_dog
Quote:

Originally Posted by william

2. is $\displaystyle sin 2x sin x- cos x = \frac{\sqrt3}{4}$ is true for $\displaystyle \frac{7\pi}{6}$? Explain

Replace x with $\displaystyle \frac{7\pi}{6}$

$\displaystyle \sin\left(2\cdot\frac{7\pi}{6}\right)=\sin\frac{7\ pi}{3}=\sin\left(2\pi+\frac{\pi}{3}\right)=\sin\fr ac{\pi}{3}=\frac{\sqrt{3}}{2}$

$\displaystyle \sin\frac{7\pi}{6}=\sin\left(\pi+\frac{\pi}{6}\rig ht)=-\sin\frac{\pi}{6}=-\frac{1}{2}$

$\displaystyle \cos\frac{7\pi}{6}=\cos\left(\pi+\frac{\pi}{6}\rig ht)=-\cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}$

I used $\displaystyle \sin(2\pi+\alpha)=\sin\alpha, \ \sin(\pi+\alpha)=-\sin\alpha, \ \cos(\pi+\alpha)=-\cos\alpha$