Maths - Please Help Me?

Question

Hello,

There are some questions that I am not sure how to tackle:

1) cot2A + cosec2A = cotA

2) sin^4θ + cos^4θ = 1/4(cos4θ + 3)

3) tan3θ = (3tanθ - tan^3θ) / (1 - 3tan^2θ)

Thank you in advance.

Answer 1

1) cot2A + cosec2A = cotA

Left hand side identity :cot2A + cosec2A

Use formulas: cotA =cosA/sinA;cosecA = 1/sinA

cot2A + cosec2A =(cos2A/sin2A)+(1/sin2A)

                         =(cos2A+1)/(sin2A)

Double angle formula : cos2A+1 =2cos²A,sin2A = 2sinAcosA

                         =(2cos²A)/(2sinAcosA)

                                  =(cosA)/(sinA)

                                   = cot A

= Right hand side identity.

 

Answer 2

3) tan3θ = (3tanθ - tan^3θ) / (1 - 3tan^2θ)

Left hand side identity :tan3θ = tan(2θ+θ)

Composite formula:tan(A+B)=(tanA+tanB)/(1-tanAtanB)

             =(tan2θ+tanθ)/(1-tan2θtanθ)

Double angle formula : tan2θ =2tanθ/(1-tan²θ)

              =((2tanθ/(1-tan²θ))+tanθ)/(1-(2tanθ/(1-tan²θ))tanθ)

               =((2tanθ)+tanθ(1-tan²θ)/(1-tan²θ)/((1-tan²θ)-(2tan²θ))/(1-tan²θ)

              =((2tanθ)+tanθ-tan³θ)/((1-tan²θ)-(2tan²θ))

              =(3tanθ-tan³θ)/(1-3tan²θ)                                 

= Right hand side identity.

Answer 3

2) sin^4θ + cos^4θ = 1/4(cos4θ + 3)

Left hand side identity :sin^4θ + cos^4θ

sin^4θ + cos^4θ =(sin²θ)² + (cos²θ)²

Double angle formula : sin²θ=(1-cos2θ)/2 and cos²θ=(1+cos2θ)/2                                  

=((1-cos2θ)/2)² + ((1+cos2θ)/2)²

=1/4((1-cos2θ)²+(1+cos2θ))²

=1/4(2(1+cos²2θ)

 =1/4(2(1+(1+cos4θ)/2)

=1/4(2(2+(1+cos4θ))/2)

=1/4(3+cos4θ)

 Right hand side identity.