
How do you simplify $\cos (\alpha + \beta ) - \cos (\alpha - \beta )$ ?
Answer
543.3k+ views
Hint: Trigonometry is that branch of mathematics that tells us the relation between the sides of a right-angled triangle that is the base, the perpendicular and the hypotenuse. All the trigonometric functions can be converted into any other trigonometric function through some identities and have great importance in mathematics and real-life too. To find the trigonometric value of large angles, there are many ways. In the given question, we will find the cosine of the sum and the difference of $\alpha $ and $\beta $ by using the sum or difference identity.
Complete step by step answer:
We have to simplify $\cos (\alpha + \beta ) - \cos (\alpha - \beta )$
We know that
$ \Rightarrow \cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $
$ \Rightarrow \cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $
Using these values in the given equation, we get –
$ \Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta - (\cos \alpha \cos \beta + \sin \alpha \sin \beta )$
We know that when $ - (a + b) = - a - b$
$ \Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta - \cos \alpha \cos \beta - \sin \alpha \sin \beta $
Writing the like terms together we get –
$
\Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = \cos \alpha \cos \beta - \cos \alpha \cos \beta - \sin \alpha \sin \beta - \sin \alpha \sin \beta \\
\Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = - 2\sin \alpha \sin \beta \\
$
Hence, the simplified form of $\cos (\alpha + \beta ) - \cos (\alpha - \beta )$ is $ - 2\sin \alpha \sin \beta $.
Note: Trigonometric functions are of six types that are sine, cosine, tangent, cosecant, secant and cotangent. Sine, cosine and tangent are the main functions while cosecant, secant and cotangent are their reciprocals respectively. According to the sum identity of cosines, the cosine of the sum of any two angles a and b is equal to the difference of the product of the cosine of one angle (a) and the cosine of the other angle (b) and the product of the sine of angle a and sine of angle b, that is, $\cos (a + b) = \cos a\cos b - \sin a\sin b$, similarly the cosine of the difference of any two angles a and b is equal to the sum of the product of the cosine of one angle (a) and the cosine of the other angle (b) and the product of the sine of angle a and sine of angle b, that is, $\cos (a - b) = \cos a\cos b + \sin a\sin b$. Using these two identities we can solve similar questions.
Complete step by step answer:
We have to simplify $\cos (\alpha + \beta ) - \cos (\alpha - \beta )$
We know that
$ \Rightarrow \cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $
$ \Rightarrow \cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $
Using these values in the given equation, we get –
$ \Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta - (\cos \alpha \cos \beta + \sin \alpha \sin \beta )$
We know that when $ - (a + b) = - a - b$
$ \Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta - \cos \alpha \cos \beta - \sin \alpha \sin \beta $
Writing the like terms together we get –
$
\Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = \cos \alpha \cos \beta - \cos \alpha \cos \beta - \sin \alpha \sin \beta - \sin \alpha \sin \beta \\
\Rightarrow \cos (\alpha + \beta ) - \cos (\alpha - \beta ) = - 2\sin \alpha \sin \beta \\
$
Hence, the simplified form of $\cos (\alpha + \beta ) - \cos (\alpha - \beta )$ is $ - 2\sin \alpha \sin \beta $.
Note: Trigonometric functions are of six types that are sine, cosine, tangent, cosecant, secant and cotangent. Sine, cosine and tangent are the main functions while cosecant, secant and cotangent are their reciprocals respectively. According to the sum identity of cosines, the cosine of the sum of any two angles a and b is equal to the difference of the product of the cosine of one angle (a) and the cosine of the other angle (b) and the product of the sine of angle a and sine of angle b, that is, $\cos (a + b) = \cos a\cos b - \sin a\sin b$, similarly the cosine of the difference of any two angles a and b is equal to the sum of the product of the cosine of one angle (a) and the cosine of the other angle (b) and the product of the sine of angle a and sine of angle b, that is, $\cos (a - b) = \cos a\cos b + \sin a\sin b$. Using these two identities we can solve similar questions.
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