
Find the value of cos75°.
Answer
541.2k+ views
Hint: Observe that 75° can be expressed as 30° + 45° and we know the values of cos45°, cos30°, sin45°, and sin30°. Use the formula \[\cos (A + B) = \cos A\cos B - \sin A\sin B\] to find the value of cos75°.
Complete step-by-step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle is called trigonometric functions.
The sine, cosine, tangent, cotangent, secant, and cosecant are the trigonometric functions.
The sine and cosecant are inverses of each other. The cosine and secant are inverses of each other. The tangent and the cotangent are inverses of each other.
They all are related to one another with special formulas.
We know the formula for the cosine of the sum of two angles is the product of the cosine of these angles minus the product of the sine of these angles. Hence, we have:
\[\cos (A + B) = \cos A\cos B - \sin A\sin B..........(1)\]
We don’t know the value of cos75° directly. But we can write 75° as the sum of 45° and 30° and use equation (1) to simplify the expression.
\[\cos (30^\circ + 45^\circ ) = \cos 30^\circ \cos 45^\circ - \sin 30^\circ \sin 45^\circ .........(2)\]
We know the value of cos30°, cos45°, sin30°, and sin45° as follows:
\[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\]
\[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\]
\[\sin 30^\circ = \dfrac{1}{2}\]
\[\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}\]
Substituting these values in the equation (2), we have:
\[\cos (30^\circ + 45^\circ ) = \dfrac{{\sqrt 3 }}{2}\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}\dfrac{1}{{\sqrt 2 }}\]
Simplifying, we get:
\[\cos (75^\circ ) = \dfrac{1}{{2\sqrt 2 }}(\sqrt 3 - 1)\]
Multiplying by \[\sqrt 2 \] to rationalize, we have:
\[\cos (75^\circ ) = \dfrac{1}{{2\sqrt 2 }}(\sqrt 3 - 1) \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\]
\[\cos (75^\circ ) = \dfrac{1}{4}(\sqrt 6 - \sqrt 2 )\]
Hence, the value of cos75° is \[\dfrac{1}{4}(\sqrt 6 - \sqrt 2 )\].
Note: The most important trick is to express 75° as the sum of 45° and 30° and then it is obvious to find the value of cos(45° + 30°). You may convert the given expression into sine and also solve for the answer.
Complete step-by-step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle is called trigonometric functions.
The sine, cosine, tangent, cotangent, secant, and cosecant are the trigonometric functions.
The sine and cosecant are inverses of each other. The cosine and secant are inverses of each other. The tangent and the cotangent are inverses of each other.
They all are related to one another with special formulas.
We know the formula for the cosine of the sum of two angles is the product of the cosine of these angles minus the product of the sine of these angles. Hence, we have:
\[\cos (A + B) = \cos A\cos B - \sin A\sin B..........(1)\]
We don’t know the value of cos75° directly. But we can write 75° as the sum of 45° and 30° and use equation (1) to simplify the expression.
\[\cos (30^\circ + 45^\circ ) = \cos 30^\circ \cos 45^\circ - \sin 30^\circ \sin 45^\circ .........(2)\]
We know the value of cos30°, cos45°, sin30°, and sin45° as follows:
\[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\]
\[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\]
\[\sin 30^\circ = \dfrac{1}{2}\]
\[\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}\]
Substituting these values in the equation (2), we have:
\[\cos (30^\circ + 45^\circ ) = \dfrac{{\sqrt 3 }}{2}\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}\dfrac{1}{{\sqrt 2 }}\]
Simplifying, we get:
\[\cos (75^\circ ) = \dfrac{1}{{2\sqrt 2 }}(\sqrt 3 - 1)\]
Multiplying by \[\sqrt 2 \] to rationalize, we have:
\[\cos (75^\circ ) = \dfrac{1}{{2\sqrt 2 }}(\sqrt 3 - 1) \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\]
\[\cos (75^\circ ) = \dfrac{1}{4}(\sqrt 6 - \sqrt 2 )\]
Hence, the value of cos75° is \[\dfrac{1}{4}(\sqrt 6 - \sqrt 2 )\].
Note: The most important trick is to express 75° as the sum of 45° and 30° and then it is obvious to find the value of cos(45° + 30°). You may convert the given expression into sine and also solve for the answer.
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