Question

Solve the expression given below, and try to find a simple expression:
$\sin 70{}^\circ -\cos 40{}^\circ$
(a). $\sin 10{}^\circ$
(b). $\cos 80{}^\circ$
(c). $\tan 10{}^\circ$
(d). $\cot 10{}^\circ$

Answer 1

Hint: When an equation is given in terms of Sine, Cosine, tangent, we must use any of the trigonometric identities to make the equation solvable. There are many inter-relations between Sine, Cosine, tan, secant. These are inter-relations called as identities. Whenever you see conditions such that $\theta \in R$ , that means inequality is true for all angles. So, directly think of identity which will make your work easy. Use: - $\tan x=\dfrac{\sin x}{\cos x},\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$ .

Complete step-by-step solution -
An equality with Sine, Cosine or tangent in them is called trigonometric equality. These are solved by some inter-relations known beforehand.
All the inter-relations which relate Sine, Cosine, tangent, Cotangent, Secant, Cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.
Given equation in the question is in terms of trigonometric terms:
$=\sin 70{}^\circ -\cos 40{}^\circ$
By basic trigonometry we know that $\cos x=\sin \left( 90-x \right)$
By using this we can write the second term as:
$=\sin 70{}^\circ -\sin 50{}^\circ$
By basic trigonometry we know the formula of $\sin A-\sin B$ given by:
$\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$
By substituting the values of A, B as 70, 50, we get the equation as:
$=2\sin \left( \dfrac{70-50}{2} \right)\cos \left( \dfrac{70+50}{2} \right)$
By simplifying the angles inside we get it as follows:
$=2\sin 10{}^\circ \cos 60{}^\circ$
By basic geometry we know the value of $\cos 60{}^\circ$ as 0.5.
By substituting this we get the expression in terms of:
$=2\sin 10{}^\circ \left( 0.5 \right)$
By simplifying we can say value of expression to be:
$\Rightarrow \sin 70{}^\circ -\cos 40{}^\circ =\sin 10{}^\circ$
$\sin 10{}^\circ$ is the value of the given expression in the question.
Therefore, option (a) is correct for the given question.

Note: Be careful while applying $\sin A-\sin B$ you must do $70-50$ generally, students confuse and write $50-70$ which is wrong. Alternate method is you can write $\sin 70$ in terms of $\cos 20$ and then apply $\cos A-\cos B$ formula though you will reach the same result, (you can try that too).