Question #1ce92

Question

421 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-07T14:10:23

#sin2a=0#

We first need to find #cosa#, so we can use the Pythagorean identity, which says that

#sin^2a+cos^2a=1#

#:.cos^2a=1-1=0#

#cosa=0#

Then, we use another trigonometric identity, which is

#sin2a=2sinacosa#

We have: #sina=1,cosa=0#

#:.sin2a=2*1*0=0#

Answer 2 · 2018-02-07T18:43:50

2cos a.

Trig identity -->
sin 2a = 2sin a.cos a
If sin a = 1 , then,
sin 2a = 2cos a.

Answer 3 · 2018-02-07T19:29:03

# 0#.

#sina=1rArr a=pi/2,5pi/2,9pi/2,...,(4n-3)pi/2, n in NN, or, #

#a=-3pi/2,-7pi/2,-11pi/2,...,-(4m-1)pi/2, m in NN#.

#"If, "a=(4n-3)pi/2, n in NN,#, then,

#sin2a=sin{2*(4n-3)pi/2}=sin{(4n-3)pi}=0#.

Similarly, if #a=-(4m-1)pi/2, m in NN#, then,

#sin2a=sin[2{-(4m-1)pi/2}]#,

#=sin{-(4m-1)pi}#,

#=-sin{(4m-1)pi}#,

#=0#.

Hence, #sin2a=0, if sina=1#.