Math Instructor

How to find values of sin 15, sin 75, cos 15, cos 75, tan 15, tan 75, cot 15, cot 75, sec 15, sec 75, cosec 15 and cosec 75 degrees?

In one of the post we learnt values of sin, cos, tan, cot, cosec and sec for 0, 30, 45, 60 and 90 degrees. You can review the post here http://mathinstructor.net/2012/08/values-of-trigonometric-ratios-for-0-30-45-60-and-90-degrees/

In this post, we will learn how can we find value of sin 15, sin 75, cos 15, cos 75, tan 15, tan 75, cot 15, cot 75, sec 15, sec 75, cosec 15 and cosec 75 degrees.

 

We know by formulas that sin(A+B) = sinA.cosB+sinB.cosA

 

sin(A-B)=sinA.cosB-sinB.cosA

 

cos (A+B) = cosA.cosB-sinA.sinB

 

cos(A-B) = cos A.cosB+sin A.sin B

 

tan(A+B) = \frac{tan A+tan B}{1-tanA.tan B}

 

tan(A-B) = \frac{tan A-tan B}{1+tanA.tan B}

 

 

sin(45-30)=sin45.cos30-sin30.cos45=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}- \frac{1}{2} \times \frac{1}{\sqrt{2}}

 

\Rightarrow sin 15=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{2\sqrt{2}}{2\sqrt{2}}=2(\frac{\sqrt{6}-\sqrt{2}}{8})=\frac{\sqrt{6}-\sqrt{2}}{4}

 

 

sin(45+30)=sin45.cos30+sin30.cos45=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+ \frac{1}{2} \times \frac{1}{\sqrt{2}}

 

\Rightarrow sin 75=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\times \frac{2\sqrt{2}}{2\sqrt{2}}=2(\frac{\sqrt{6}+\sqrt{2}}{8})=\frac{\sqrt{6}+\sqrt{2}}{4}

 

 

 

Similarly, we can find value of cos (A+B), cos (A-B), tan (A+B) and tan (A-B).

 

Also, cot(A+B)=\frac{1}{tan(A+B)}, sec (A+B) = \frac{1}{cos(A+B)} and cosec (A+B)=\frac{1}{sin(A+B)}


 


Posted in : Trigonometry


Tags :
No Comments




Leave a Reply

Your email address will not be published. Required fields are marked *