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Question: The value of \[\left( \dfrac{\sin 55-\cos 55}{\sin 10} \right)\] is equal to? \[\begin{aligned} ...

The value of (sin⁡55−cos⁡55sin⁡10)\left( \dfrac{\sin 55-\cos 55}{\sin 10} \right) is equal to?

& 1.\dfrac{1}{\sqrt{2}} \\\ & 2.2 \\\ & 3.1 \\\ & 4.\sqrt{2} \\\ \end{aligned}$$
Explanation

Solution

In order to find the value of (sin⁡55−cos⁡55sin⁡10)\left( \dfrac{\sin 55-\cos 55}{\sin 10} \right), we will be expressing sin⁡55\sin 55 angle in terms of principle angle. Then we obtain both the angles in the braces in terms of cos. Then we will be solving it by applying the formula cos⁡x−cos⁡y\cos x-\cos y. Solving it accordingly will give us the required answer.

Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between 0∘{{0}^{\circ }} and 360∘{{360}^{\circ }}. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us find the value of (sin⁡55−cos⁡55sin⁡10)\left( \dfrac{\sin 55-\cos 55}{\sin 10} \right)
Firstly, let us express sin⁡55\sin 55 in terms of cosine.
⇒(sin⁡55−cos⁡55sin⁡10)=sin⁡(90−35)−cos⁡55sin⁡10\Rightarrow \left( \dfrac{\sin 55-\cos 55}{\sin 10} \right)=\dfrac{\sin \left( 90-35 \right)-\cos 55}{\sin 10}
⇒cos⁡35−cos⁡55sin⁡10\Rightarrow \dfrac{\cos 35-\cos 55}{\sin 10} because sin⁡90−θ=cos⁡θ\sin 90-\theta =\cos \theta
Now we will be applying the formula cos⁡x−cos⁡y=−2sin⁡x+y2sin⁡x−y2\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}
We get,
⇒−2sin⁡35+552sin⁡35−552sin⁡10\Rightarrow \dfrac{-2\sin \dfrac{35+55}{2}\sin \dfrac{35-55}{2}}{\sin 10}
Upon solving this, we get
⇒−2sin⁡45(−sin⁡10)sin⁡10=2sin⁡45\Rightarrow \dfrac{-2\sin 45\left( -\sin 10 \right)}{\sin 10}=2\sin 45
We know that sin⁡45=12\sin 45=\dfrac{1}{\sqrt{2}}
Now let us substitute the value and solve it. We get
⇒2×12=2\Rightarrow 2\times \dfrac{1}{\sqrt{2}}=\sqrt{2}

Note: While solving problems, we must try to express the angles in terms of principal angles. We must always be aware of the trigonometric values for substitution and solving. We must not forget to check out for the general formulas that relate to the obtained equation for our easy simplification of trigonometric functions.