Question

if cosA =3/5,then find cos2A,sin2A,cos3A?

Answer 1

If...
cosA = 3/5 then sinA = 4/5

cos 2A = cos²A - sin ² A
= (3/5)²- (4/5)²
=9/5 - 16/5
=(9-16)/5
=-7/5

sin 2A = 2sinAcosA
=2×4/5×3/5
=24/25

cos 3A = 4 cos³A - 3 cos A
=4(3/5)³-3(3/5)
=4(27/125) - 9/5
=108/125-9/5
=(108-27)/125
=81/125
I hope it helps u

Answer 2

Answer:

  cos 2A  =  -7/25 ,  sin 2A = 24/25   cos 3A = - 117/125  

Explanation:

Given problem

cos A = 3/5  then find  i)  cos2A,  ii)  sin2A,  iii)  cos3A  

⇒ given that cosA = $\frac{3}{5}$  

we know that in a right angled triangle  cos θ =  $\frac{adjacent side }{hypotenuse}$  

⇒ adjacent side = 3    and    hypotenuse = 5

⇒ from Pythagorean theorem in a right angled triangle

     hypotenuse² = adjacent side² + opposite side²

⇒   5²  =  3² + opposite side²

⇒ opposite side² = 5² - 3²  

                             =  25 - 9 = 16 = 4²      

     opposite side =  4  

⇒  sin A = $\frac{opposite side}{hypotenuse} = \frac{4}{5}$    

i)  cos 2A  

⇒ cos 2A = cos²A - sin²A

                = $[\frac{3}{5}]^{2} - [\frac{4}{5} ] ^{2}$ =  $\frac{9}{25} - \frac{16}{25}$  $=-\frac{7}{25}$  

ii)  sin 2A  

⇒ sin 2A =  2 sin A cos A

               =   $2(\frac{4}{5} )(\frac{3}{5})$  = $\frac{24}{25}$      

 iii)  cos 3A  

 ⇒ cos 3A =  4 cos³A - 3 cos A

                  = $4(\frac{3}{5})^{3}-3(\frac{3}{5} )$  

                  = $4(\frac{27}{125} ) - 3(\frac{3}{5} )$  

                  = $\frac{108}{125} - \frac{9}{5}$  = $\frac{108 - 225}{125}$ =  $-\frac{117}{125}$