Question

If we have the trigonometric function as x = $\sin 130^\circ \cos 80^\circ $, y = $\sin 80^\circ \cos 130^\circ $, z = 1 + xy which one of the following is true?
1). X>0, y>0, z>0
2). X>0, y<0, 0< z <1
3). X>0, y<0, z>1
4). X<0, y<0, 0< z <1

Answer 1

Hint: we have to solve the first and second equation to find the value of x and y. Then, the value of x and y is to be put in the third equation (z = 1+xy) and by this we will get to know the value of z. All these three values should be compared from the options given and we will get the correct one.

Complete step-by-step solution:
Given: x = $\sin 130^\circ \cos 80^\circ $-----(1)
Y = $\sin 80^\circ \cos 130^\circ $--------(2)
Z = 1 + xy--------(3)
Now, first we will solve equation (1)
$x = \sin 130^\circ \cos 80^\circ $
Now, we will divide and multiply the equation with 2
$x = (\sin 130^\circ \cos 80^\circ )\dfrac{2}{2}$
$\Rightarrow x = \dfrac{{2\sin 130^\circ \cos 80^\circ }}{2}$
By using the property $2\sin A \cos B = \sin(A+B)+\sin(A-B)$, we will rewrite this equation
$x = \dfrac{1}{2}\left( {\sin 210^\circ + \sin 50^\circ } \right)$
\[\Rightarrow x = \dfrac{1}{2}\left( {\sin \left( {270^\circ - 60^\circ } \right) + \sin 50^\circ } \right)\]
$\Rightarrow x = \dfrac{1}{2}\left( { - \cos 60^\circ + \sin 50^\circ } \right)$
The value of $\cos 60^\circ $is 1/2. So, we will replace it in this equation.
$x = \dfrac{1}{2}\left( {\sin 50^\circ - \dfrac{1}{2}} \right)$
$\Rightarrow x = \dfrac{1}{2}\left( {\sin 50^\circ - \sin 30^\circ } \right)$
Sine is an increasing function. So, $\sin 50^\circ $ is greater than $\sin 30^\circ $.
$\sin 50^\circ > \sin 30^\circ $
Then, $\dfrac{{\sin 50^\circ - \sin 30^\circ }}{2} > 0$
So, \[x > 0\]
Now, we will solve equation (2)
$y = \sin 80^\circ \cos 130^\circ $
Now, same as above we will divide and multiply this equation also with 2.
$y = \left( {\sin 80^\circ \cos 130^\circ } \right) \times \dfrac{2}{2}$
$\Rightarrow y = \dfrac{{2\cos 130^\circ \sin 80^\circ }}{2}$
Now, by using the property $ 2\cos A \sin B =\sin(A+B)-\sin(A-B)$, we will rewrite this equation.
$y = \dfrac{1}{2}\left( {\sin 210^\circ - \sin 50^\circ } \right)$
$\Rightarrow y = \dfrac{1}{2}\left( { - \cos 60^\circ - \sin 50^\circ } \right)$
$\Rightarrow y = - \dfrac{1}{2}\left( {\sin 50^\circ + \dfrac{1}{2}} \right)$
$\Rightarrow y = - \dfrac{1}{2}\left( {\sin 50^\circ + \sin 30^\circ } \right)$
Sine is an increasing function. So,
$\sin 50^\circ + \sin 30^\circ > 0$
$ - \dfrac{1}{2}\left( {\sin 50^\circ + \sin 30^\circ } \right) < 0$
So, $y < 0$
Now, we will solve equation (3)
$z = 1 + xy$
$\Rightarrow z = 1 + \left( {\dfrac{{\sin 50^\circ - \sin 30^\circ }}{2}} \right)\left( { - \dfrac{{\left( {\sin 50^\circ + \sin 30^\circ } \right)}}{2}} \right)$
$z = 1 - \dfrac{1}{4}\left( {{{\sin }^2}50^\circ - {{\sin }^2}30^\circ } \right)$
Value of $\sin 30^\circ $is 1/2. So, we will replace it in this equation.
$z = 1 - \dfrac{1}{4}\left( {{{\sin }^2}50 - \dfrac{1}{4}} \right)$
$\Rightarrow z = 1 - \dfrac{{{{\sin }^2}50^\circ }}{4} + \dfrac{1}{{16}}$
$\Rightarrow z = \dfrac{{17}}{{16}} - \dfrac{{{{\sin }^2}50}}{4}$
Value of ${\sin ^2}50^\circ $is greater than 1/4 but less than 3/4.
So, $\dfrac{1}{4} < {\sin ^2}50^\circ < \dfrac{3}{4}$
$\dfrac{{17}}{{16}} - \dfrac{{{{\sin }^2}50}}{4} > 0$
$z > 0$ and $z < 1$
Hence, x>0, y<0, 0< z< 1. So, option 2. Is the correct answer.

Note: Properties of trigonometric functions are used to solve these kinds of questions. Values of trigonometric angles should be remembered while solving the question. Equations must be solved step by step and calculations are to be performed very carefully.