Trigonometry> find the value of sin20degree*sin40degree...

find the value of sin20degreesin40degreesin80degree =

keerthi , 8 Years ago

Grade 11

3 Answers

Lucky

For solving such type of questions you can refer to r.d Sharma of 11 class and chapter 8 (transformation formula) the solution is way too long but I get my answer = square root of 3 divided by 8.

Last Activity: 8 Years ago

Arun

I'm assuming that you know this:
sin(A+B) = sin A cos B + cos A sin B

Step number 1:
write sin(40) as sin(60-20)
write sin(80) as sin(60+20)
since 60 degrees is a remarkable angle, you know that sin(60) = √(3)/2

Step 2:
you have this:
sin(20)•sin(60-20)•sin(60+20)

- Simplify sin(60-20)•sin(60+20)
sin(60-20) = sin60•cos20 - sin20•cos60
sin(60+20)= sin60•cos20 + sin20•cos60
- multiply them:
= sin²60•cos²20 - sin²20•cos²60

Step 3:
- you know that sin²A+cos²A=1,
- so, cos²A = 1 - sin²A

- Simplify sin²60•cos²20 - sin²20•cos²60
= sin²60•(1-sin²20) - sin²20•(1-sin²60)
= sin²60- sin²20•sin²60 - sin²20 + sin²20•sin²60
= sin²60 - sin²20
= 3/4 - sin²20

Step 4:
This is what we have now:
= sin20•(3/4 - sin²20)
= [1/4]•(3sin20 - 4sin³20)
- note that 4sin³20 comes from putting [(3sin20)/4]-sin³20 on the same base

Step 5:
you need to know that
sin(3A) = 3sin(A) - 4sin³(A)
(multiple angle relations)

- so, (3sin20 - 4sin³20) = sin60

Step 6:
now you have this:
= [1/4]•sin60
= [1/4]•[√(3)/2)
=√(3)/8

Regards

Arun (askIITians forum expert)

Last Activity: 8 Years ago