# Trigonometry multiplication?

How would I solve the following problem?

$$\cos(67.5^\circ) \cos(22.5^\circ)$$

I multiplied them using wolfram alpha and got $.353553$ but how would I find an exact value?

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yes you are correct. –  Fernando Martinez Feb 5 at 2:42

$$\cos(x^{\circ}) \cos((90-x)^{\circ}) = \cos(x^{\circ}) \sin(x^{\circ}) = \dfrac{\sin((2x)^{\circ})}2$$ In your case, take $x^{\circ} = 22.5^{\circ}$.

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yes yes this is the answer I have been searching for. –  Fernando Martinez Feb 5 at 2:48

Hint:

$$\cos x+\cos y=2\;\cos\frac{x+y}{2}\cdot\cos\frac{x-y}{2}$$

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I see a sum to product formula. –  Fernando Martinez Feb 5 at 2:38
So I would add $67.5$ plus $22.5$ divide it by $2$ and then multiply it by the right side and then multiply by $2$ but when I add them I got $89.75$ do I round to $90$ –  Fernando Martinez Feb 5 at 2:41
+1 Straightforward answer. –  B.S. Feb 5 at 2:55

Observe that $67.5 = 45 + 22.5$ and $22.5 = 45 - 22.5$. Combine this with the other answers provided, and you should be on your way...consider substituting $x = 45$ and $y=22.5$

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$$2\cos A\cos B=\cos(A-B)+\cos(A+B)$$

So, $$2\cos 67.5^\circ\cos22.5^\circ=\cos(67.5^\circ-22.5^\circ)+\cos(67.5^\circ+22.5^\circ)=\cos45^\circ+\cos90^\circ=\frac1{\sqrt2}+0$$

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lets use half angle formulas we know $cos^2\theta = \frac{1 + cos2\theta}{2}$ now if we just plug in $cos(22.5^{\circ})$ and $cos(66.5^{\circ})$ we would get, i will do $cos(22.5^{\circ})$ first and you could do $cos(66.5^{\circ})$ it would would work the same way alright
so $cos^2(22.5^{\circ}) = \frac{1 + cos2\cdot22.5^{\circ}}{2} \Rightarrow cos^2(22.5^{\circ}) = \frac{1 + cos45^{\circ}}{2} \Rightarrow cos^2(22.5^{\circ}) = \frac{1 + \frac{1}{\sqrt{2}}}{2} \Rightarrow cos^2(22.5^{\circ}) = \frac{\sqrt{2} + 1}{2\sqrt{2}} = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} \Rightarrow \sqrt{\frac{\sqrt{2} + 2}{4}} = cos(22.5^{\circ})$
in exact form now do $cos(66.5^{\circ})$ in similar fasion and multiply them together :)