Home › Forums › Math questions › Trigonometry questions

Find the value of :

Posted October 26th, 2009 by alpha
in

$ 2(\cos^2 76^0+\cos^216^0-\cos76^0\cos16^0)=1+\cos 152^o+1+\cos32^o-\cos 60^o-\cos 92^o=\frac{3}{2}+\cos 32^o-\cos 28^o+\sin 2^o=\frac{3}{2}-2\sin 30^o\sin 2^o+\sin 2^o=\frac{3}{2} $
structure

‹ Prove that Prove that triangle is either an isosceles or a right angled triangle. ›

Comments

Formula

On October 26th, 2009 alpha says:

Which formula is to be used?

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On October 26th, 2009 Structure says:

$ 2(\cos^2 76^0+\cos^216^0-\cos76^0\cos16^0)=1+\cos 152^o+1+\cos32^o-\cos 60^o-\cos 92^o=\frac{3}{2}+\cos 32^o-\cos 28^o+\sin 2^o=\frac{3}{2}-2\sin 30^o\sin 2^o+\sin 2^o=\frac{3}{2} $

The answer is 3/4

On October 28th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt
You were not attentive. You are right, the answer is $ \frac{3}{4} $ but I have evaluated the double of your expression to avoid to write fractions.

In fact your exercise is a particular case of
more general

$$2cos^2(\frac{\pi}{3}+x)+2\cos^2x-2\cos(\frac{\pi}{3}+x)\cos x=\frac{3}{2}$$

as

$$1+\cos(\frac{2\pi}{3}+2x)+1+\cos 2x-\cos(\frac{\pi}{3}+2x)-\cos \frac{\pi}{3}=\frac{3}{2}+2\cos(\frac{\pi}{3}+2x)\cos\frac{\pi}{3}-\cos(\frac{\pi}{3}+2x)=\frac{3}{2}$$

Your

$$x=16^o$$

in degrees.
Best regards
''ALPHA" =)

Back to top