I am having problems with proving identities that has a number and an exponent with it.
Like:
*2cosxcsc2x = cscx
*sin^2 2x = 4sin^2 x cos^2 x
*cot2(theta)=csc(theta)-2sin(theta)/2cos(theta)
2cosxcsc2x = cscx
LS = 2cosx(1/sin 2x)
= 2cosx(1/(2sinxcosx)
= 1/sinx
= csc x
= RS
sin^2 2x = 4 sin^2 x cos^2 x
will use the same identity as in the first one:
namely, sin 2x = 2sinxcosx
LS = sin^2 2x
= (sin 2x)(sin 2x)
= (2sinxcosx)(2sinxcosx)
= 4 sin^2 x cos^2 x
= RS
cot 2Ø = cscØ - 2sinØ/2cosØ
I think it should be
cot 2Ø = (cscØ - 2sinØ)/2cosØ
RS = (1/sinØ - 2sinØ)/2cosØ
multiply top and bottom by sinØ
= (1 - 2sin^2 Ø)/(2sinØcosØ ----> recall cos 2x - 1 - 2sin^2 x
= cos 2Ø / sin 2Ø
= cot 2Ø
= LS
When dealing with proving trigonometric identities involving numbers and exponents, it's important to have a good understanding of the trigonometric functions and their properties. Here are step-by-step explanations for proving each of the identities you mentioned:
1. Proving 2cos(x)csc(2x) = csc(x):
To prove this identity, start by expressing all the trigonometric functions in terms of sine and cosine.
csc x = 1/sin x
csc 2x = 1/sin 2x = 1/2sin x cos x (using double angle identity)
Now substitute these expressions into the original equation:
2cos x * (1/2sin x cos x) = 1/sin x
Simplify the left side by canceling the common factors:
cos x * cos x = 1 (since 2 * 1/2 = 1)
This is a true statement, so the identity is proved.
2. Proving sin^2(2x) = 4sin^2(x)cos^2(x):
To prove this identity, start by using the double angle identity for sine:
sin(2x) = 2sin(x)cos(x)
Now substitute this expression into the original equation:
(2sin(x)cos(x))^2 = 4sin^2(x)cos^2(x)
Expand and simplify the left side:
4sin^2(x)cos^2(x) = 4sin^2(x)cos^2(x)
This is a true statement, so the identity is proved.
3. Proving cot(2θ) = csc(θ) - 2sin(θ) / 2cos(θ):
To prove this identity, start by expressing the trigonometric functions in terms of sine and cosine:
cot(2θ) = cos(2θ)/sin(2θ)
csc(θ) = 1/sin(θ)
sin(θ) = 2sin(θ)cos(θ) (using double angle identity)
Now substitute these expressions into the original equation:
cos(2θ) / sin(2θ) = 1/sin(θ) - 2sin(θ) / 2cos(θ)
Multiply both sides by sin(2θ) to clear the fractions:
cos(2θ) = 2 - 2sin^2(θ) / cos(θ)
Rewrite sin^2(θ) using the Pythagorean identity:
cos(2θ) = 2 - 2(1 - cos^2(θ)) / cos(θ)
Distribute and combine like terms:
cos(2θ) = 2cos^2(θ) / cos(θ)
Simplify:
cos(2θ) = 2cos(θ)
This is a true statement, so the identity is proved.
Remember, when proving trigonometric identities, it's important to manipulate the expressions using known trigonometric identities and algebraic techniques until both sides of the equation are equal.