Answer to Question #144318 in Trigonometry for Shahir Sheikh

Question

cos(15°) can be evaluated using a sum or difference trig identity. Which of the following is equivalent to cos(15°)?

Answer 1

Let's apply the cos of a sum formula to $\cos(15^\circ+15^\circ)$, as we know that $\cos(30^{\circ})=\frac{\sqrt{3}}{2}:$

$\frac{\sqrt{3}}{2} = \cos(30^\circ)=\cos(15^\circ+15^\circ)=\cos(15^\circ)\times\cos(15^\circ)-\sin(15^\circ)\times\sin(15^\circ)$

Now by the main trigonometrical identity ($\sin^2\alpha+\cos^2\alpha=1)$ we can write :

$\frac{\sqrt{3}}{2}=\cos^2(15^\circ)-(1-\cos^2(15^\circ))=2\cos^2(15^\circ)-1$

$2\cos^2(15^\circ)=1+\frac{\sqrt{3}}{2}$

$\cos(15^\circ)=\pm \sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}$

We know also that $\cos(15^\circ)>0$ (as $0^\circ<15^\circ<90^\circ$ ), so we have :

$\cos(15^\circ)=\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}=\frac{1}{2} \sqrt{2+\sqrt{3}}$

We can also simplify the last expression by writing :

$(x+y)^2=2+\sqrt{3}$

$x^2+y^2=2, xy=\frac{\sqrt{3}}{2}$

$x^2+\frac{3}{4x^2}=2$

$x^4-2x^2+\frac{3}{4}=0$

$x^2=\frac{2\pm\sqrt{4-3}}{2}$

$x=\sqrt{\frac{1}{2}},y=\sqrt{\frac{3}{2}}$

So we have :

$\cos(15^\circ)=\frac{1}{2}\times(\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}})=\frac{\sqrt{2}+\sqrt{6}}{4}$