Solveeit Logo
Login

Question

Question: Find the value of \(\sin {{120}^{0}}\cos {{150}^{0}}-\cos {{240}^{0}}\sin {{330}^{0}}\) . (a) \(1\...

Find the value of sin⁡1200cos⁡1500−cos⁡2400sin⁡3300\sin {{120}^{0}}\cos {{150}^{0}}-\cos {{240}^{0}}\sin {{330}^{0}} .
(a) 11
(b) −1-1
(c) 23\dfrac{2}{3}
(d) −(3+14)-\left( \dfrac{\sqrt{3}+1}{4} \right)

Explanation

Solution

Hint: For solving this problem we will use certain formulas and then put the correct values in the given expression and choose the correct option.

Complete step by step answer:
Given:
We have to calculate the value of sin⁡1200cos⁡1500−cos⁡2400sin⁡3300\sin {{120}^{0}}\cos {{150}^{0}}-\cos {{240}^{0}}\sin {{330}^{0}} .
We will use the following formulas to solve this question:
sin⁡300=12.............(1) cos⁡300=32..........(2) cos⁡600=12.............(3) sin⁡(900+θ)=cos⁡θ..........(4) cos⁡(1800−θ)=−cos⁡θ...........(5) cos⁡(1800+θ)=−cos⁡θ............(6) sin⁡(3600−θ)=−sin⁡θ.............(7) \begin{aligned} & \sin {{30}^{0}}=\dfrac{1}{2}.............\left( 1 \right) \\\ & \cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}..........\left( 2 \right) \\\ & \cos {{60}^{0}}=\dfrac{1}{2}.............\left( 3 \right) \\\ & \sin \left( {{90}^{0}}+\theta \right)=\cos \theta ..........\left( 4 \right) \\\ & \cos \left( {{180}^{0}}-\theta \right)=-\cos \theta ...........\left( 5 \right) \\\ & \cos \left( {{180}^{0}}+\theta \right)=-\cos \theta ............\left( 6 \right) \\\ & \sin \left( {{360}^{0}}-\theta \right)=-\sin \theta .............\left( 7 \right) \\\ \end{aligned}
Now, we will use the above formulas to solve this question.
First, we will find the value of sin⁡1200\sin {{120}^{0}} , cos⁡1500\cos {{150}^{0}} , cos⁡2400\cos {{240}^{0}} and sin⁡3300\sin {{330}^{0}} .
Now, we can write, sin⁡1200=sin⁡(900+300)\sin {{120}^{0}}=\sin \left( {{90}^{0}}+{{30}^{0}} \right) so, using (4) and (2). Then,
sin⁡1200=cos⁡300=32.............(8)\sin {{120}^{0}}=\cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}.............\left( 8 \right)
Now, we can write, cos⁡1500=cos⁡(1800−300)\cos {{150}^{0}}=\cos \left( {{180}^{0}}-{{30}^{0}} \right) so, using (5) and (2). Then,
cos⁡1500=−cos⁡300=−32...............(9)\cos {{150}^{0}}=-\cos {{30}^{0}}=-\dfrac{\sqrt{3}}{2}...............\left( 9 \right)
Now, we can write, cos⁡2400=cos⁡(1800+600)\cos {{240}^{0}}=\cos \left( {{180}^{0}}+{{60}^{0}} \right) so, using (6) and (3). Then,
cos⁡2400=−cos⁡600=−12............(10)\cos {{240}^{0}}=-\cos {{60}^{0}}=-\dfrac{1}{2}............(10)
Now, we can write, sin⁡3300=sin⁡(3600−300)\sin {{330}^{0}}=\sin \left( {{360}^{0}}-{{30}^{0}} \right) so, using (7) and (1). Then,
sin⁡3300=−sin⁡300=−12.............(11)\sin {{330}^{0}}=-\sin {{30}^{0}}=-\dfrac{1}{2}.............\left( 11 \right)
Now, we can evaluate the value of sin⁡1200cos⁡1500−cos⁡2400sin⁡3300\sin {{120}^{0}}\cos {{150}^{0}}-\cos {{240}^{0}}\sin {{330}^{0}} by substituting the value of each term from equation (8), (9), (10) and (11). Then,
sin⁡1200cos⁡1500−cos⁡2400sin⁡3300 ⇒32×(−32)−(−12)×(−12) ⇒−34−12 ⇒−1 \begin{aligned} & \sin {{120}^{0}}\cos {{150}^{0}}-\cos {{240}^{0}}\sin {{330}^{0}} \\\ & \Rightarrow \dfrac{\sqrt{3}}{2}\times \left( -\dfrac{\sqrt{3}}{2} \right)-\left( -\dfrac{1}{2} \right)\times \left( -\dfrac{1}{2} \right) \\\ & \Rightarrow -\dfrac{3}{4}-\dfrac{1}{2} \\\ & \Rightarrow -1 \\\ \end{aligned}
Thus, finally, we can write that, sin⁡1200cos⁡1500−cos⁡2400sin⁡3300=−1\sin {{120}^{0}}\cos {{150}^{0}}-\cos {{240}^{0}}\sin {{330}^{0}}=-1 .
Hence, (b) is the correct option.

Note: Although the problem is very easy to solve but here, the student must take care of signs of the values while doing substitution for each term and we should use the formulae correctly and should also do the calculations without any mistake. So, we get the correct answer.