I wish to prove cos^2x-sin^2x=2cos^2x-1.
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calendarEducator since 2013
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Alternatively, we can start with the right-hand side and show that it equals the left hand side.
Rewrite the right hand side, 2(cosx)² - 1, as (cosx)² + (cosx)² - 1, which can also be rewritten as
(cosx)² - (1 - (cosx)²) by switching the order of the second and third terms and factoring out negative sign.
Then, according to Pythagorean Identity, (sinx)² + (cosx)² = 1, the expression in the parenthesis, 1 - (cosx)², can be replaced by (sinx)²:
(cosx)² - (1 - (cosx)²) = (cosx)² - (sinx)².
This, we have shown that (cosx)² - (sinx)² = 2(cosx)² - 1.
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calendarEducator since 2017
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Another way of solving the problem (cos x)² - (sin x)² = 2(cos x)² - 1
is as follows:
Consider the right-hand side (RHS) of the equation: 2(cos x)² - 1
From the Pythagorean Identity, we know that (cos x)² + (sin x)² = 1
Thus substitute 1 with (cos x)² + (sin x)² in the RHS of the equation:
2 (cos x)² - ((sin x)² + (cos x)² ).
Expand to give 2(cos x)² - (sin x)² - (cos x)² = (cos x)² - (sin x)² (as given by the LHS of the original equation).
Thus (cos x)² - (sin x)² = 2 (cos x)² - 1
calendarEducator since 2010
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We have to prove that: (cos x)^2 - (sin x)^2 = 2*(cos x)^2 - 1
Starting with the left hand side:
We use the relation (cos x)^2 + (sin x)^2 = 1 or (sin x)^2 = 1 - (cos x)^2
=> (cos x)^2 - (sin x)^2
=> (cos x)^2 - (1 - (cos x)^2)
=> (cos x)^2 - 1 + (cos x)^2)
=> 2(cos x)^2 - 1
which is the right hand side.
Therefore we prove that (cos x)^2 - (sin x)^2 = 2*(cos x)^2 - 1
You can manipulate only the left side of the equation.
For this reason, you'll use the Pythagorean identity:
(sin x)^2 + (cos x)^2 = 1
(sin x)^2 = 1 - (cos x)^2
You'll substitute (sin x)^2 to the left side of the equation:
(cos x)^2 - [1 - (cos x)^2] = 2 (cos x)^2 - 1
We'll remove the brackets to the left side:
(cos x)^2 - 1 + (cos x)^2 = 2 (cos x)^2 - 1
We'll combine like terms and we notice that LHS = RHS:
2 (cos x)^2 - 1 = 2 (cos x)^2 - 1
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