Trigonometry help plz?

Question

1.show that the equation

sin (x - 60' ) - cos ( 30' - x ) = 1


can be written in the form cos x = k, where k is a constant

2.Prove that sin^2 @ ( cot^2 @ - tan^2 @ ) = 4 cos 2@

@ represents pheta

^ represents to the power of.

Answer 1

One problem per question please.

2. Prove that sin²θ (cot²θ - tan²θ) = 4 cos 2θ

I think you meant "theta".

sin²θ (cot²θ - tan²θ) = cos²θ - sin^4 θ / cos²θ

= cos^4 θ / cos²θ - sin^4 θ / cos²θ

= (cos^4 θ - sin^4 θ) / cos²θ

= (cos²θ + sin²θ)(cos²θ - sin²θ) / cos²θ

= 1*(cos²θ - sin²θ) / cos²θ = (cos²θ - sin²θ) / cos²θ

= cos 2θ / cos²θ = cos 2θ / [(1 + cos 2θ)/2]

= 2cos 2θ / (1 + cos 2θ)

= 2cos 2θ / (cos 2θ + 1) ≠ 4 cos 2θ


This is NOT an identity. It is not true for all θ.

Answer 2

1. sin (x-60)-cos(30-x)=1

ur question is rong because the graph doesnot support the equation.......sin and cos move opposite to each oder and their differnce can never be 1...
2. sin^2@(cot^2@-tan^2@)=4cos2@

if ur preparing for sm competetion......assume any value of @........say 45
LHS
=sin^2(30)(cot^2(30)-tan^2(30))
=1/2(1-1)
=0

RHS
= 4cos2@
= 4 cos90
= 0
if u r in sm class and needs subjective explanation then i think the person at the tops have answered ur ques

Answer 3

1) sin (x - 60' ) - cos ( 30' - x ) = 1

i.e -cos (90+x - 60' ) - cos ( 30' - x ) = 1

i.e -cos (30+x ) - cos ( 30' - x )=1

i.e 2cos 30 cos x = -1

i.e . cos x = -1 / (root 3)

so k = -1 / (root 3)

2) i think the question was wrong .. i got the answer as

(2cos2@ ) /( 2cos2@+1)

Answer 4

One problem per question please.

2. Prove that sin²θ (cot²θ - tan²θ) = 4 cos 2θ

I think you meant "theta".

sin²θ (cot²θ - tan²θ) = cos²θ - sin^4 θ / cos²θ

= cos^4 θ / cos²θ - sin^4 θ / cos²θ

= (cos^4 θ - sin^4 θ) / cos²θ

= (cos²θ + sin²θ)(cos²θ - sin²θ) / cos²θ

= 1*(cos²θ - sin²θ) / cos²θ = (cos²θ - sin²θ) / cos²θ

= cos 2θ / cos²θ = cos 2θ / [(1 + cos 2θ)/2]

= 2cos 2θ / (1 + cos 2θ)

= 2cos 2θ / (cos 2θ + 1) ≠ 4 cos 2θ


This is NOT an identity. It is not true for all θ.

Answer 5

sin