Answer to Question #144318 in Trigonometry for Shahir Sheikh

Question #144318
cos(15°) can be evaluated using a sum or difference trig identity. Which of the following is equivalent to cos(15°)?
1
Expert's answer
2020-11-16T14:03:06-0500

Let's apply the cos of a sum formula to cos(15+15)\cos(15^\circ+15^\circ), as we know that cos(30)=32:\cos(30^{\circ})=\frac{\sqrt{3}}{2}:

32=cos(30)=cos(15+15)=cos(15)×cos(15)sin(15)×sin(15)\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 (sin2α+cos2α=1)\sin^2\alpha+\cos^2\alpha=1) we can write :

32=cos2(15)(1cos2(15))=2cos2(15)1\frac{\sqrt{3}}{2}=\cos^2(15^\circ)-(1-\cos^2(15^\circ))=2\cos^2(15^\circ)-1

2cos2(15)=1+322\cos^2(15^\circ)=1+\frac{\sqrt{3}}{2}

cos(15)=±12+34\cos(15^\circ)=\pm \sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}

We know also that cos(15)>0\cos(15^\circ)>0 (as 0<15<900^\circ<15^\circ<90^\circ ), so we have :

cos(15)=12+34=122+3\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+3(x+y)^2=2+\sqrt{3}

x2+y2=2,xy=32x^2+y^2=2, xy=\frac{\sqrt{3}}{2}

x2+34x2=2x^2+\frac{3}{4x^2}=2

x42x2+34=0x^4-2x^2+\frac{3}{4}=0

x2=2±432x^2=\frac{2\pm\sqrt{4-3}}{2}

x=12,y=32x=\sqrt{\frac{1}{2}},y=\sqrt{\frac{3}{2}}

So we have :

cos(15)=12×(12+32)=2+64\cos(15^\circ)=\frac{1}{2}\times(\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}})=\frac{\sqrt{2}+\sqrt{6}}{4}




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