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Question

Question: Find the value of the expression given below, \(\sin 30{}^\circ \cos 60{}^\circ +\sin 60{}^\circ \...

Find the value of the expression given below,
sin30cos60+sin60cos30\sin 30{}^\circ \cos 60{}^\circ +\sin 60{}^\circ \cos 30{}^\circ
[a] 12\dfrac{1}{2}
[b] 32\dfrac{\sqrt{3}}{2}
[c] 1
[d] 14\dfrac{1}{4}

Explanation

Solution

Substitute the values of sin30,sin60,cos30,cos60\sin 30{}^\circ ,\sin 60{}^\circ ,\cos 30{}^\circ ,\cos 60{}^\circ and simplify and hence find the value of the given expression

Complete step-by-step answer:
To solve the given question, we need to remember the values of sin30,sin60,cos30\sin 30{}^\circ ,\sin 60{}^\circ ,\cos 30{}^\circ and cos60\cos 60{}^\circ
Consider the following tables of values of sine, cosine, tangent, cotangent, secant and cosecant for angles of measure 0,30,45,60,900{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ ,90{}^\circ

From the above table, we have
sin30=12,sin60=32,cos30=32,cos60=12\sin 30{}^\circ =\dfrac{1}{2},\sin 60{}^\circ =\dfrac{\sqrt{3}}{2},\cos 30{}^\circ =\dfrac{\sqrt{3}}{2},\cos 60{}^\circ =\dfrac{1}{2}
Substituting the values of sin30,sin60,cos30,cos60\sin 30{}^\circ ,\sin 60{}^\circ ,\cos 30{}^\circ ,\cos 60{}^\circ , we get
S=12×12+32×32S=\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}
Simplifying, we get
S=14+34=44=1S=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1
Hence, we have
sin30cos60+sin60cos30\sin 30{}^\circ \cos 60{}^\circ +\sin 60{}^\circ \cos 30{}^\circ is equal to 1.
Therefore the correct answer is option (c).

Note: [1] Alternatively, you can use the fact that cos(90θ)=sinθ\cos \left( 90{}^\circ -\theta \right)=\sin \theta and sin(90θ)=cosθ\sin \left( 90{}^\circ -\theta \right)=\cos \theta and then write cos60\cos 60{}^\circ as cos(9030)=sin30\cos \left( 90{}^\circ -30{}^\circ \right)=\sin 30{}^\circ and sin60\sin 60{}^\circ as sin(9030)=cos30\sin \left( 90{}^\circ -30{}^\circ \right)=\cos 30{}^\circ
Hence the expression becomes
S=sin230+cos230S={{\sin }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ
Now, we know that sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Hence, we have
S=1S=1, which is the same as obtained above.
Hence option [c] is the correct answer.
[2] Alternative
We know that sinAcosB+cosAsinB=sin(A+B)\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)
Put A=30A=30{}^\circ and B=60B=60{}^\circ .
Hence, we have
sin(30+60)=sin30cos60+cos30sin60=S\sin \left( 30{}^\circ +60{}^\circ \right)=\sin 30{}^\circ \cos 60{}^\circ +\cos 30{}^\circ \sin 60{}^\circ =S
Hence, we have
S=sin90S=\sin 90{}^\circ
From the above table, we have sin90=1\sin 90{}^\circ =1
Hence, we have S=1S=1, which is the same as obtained above.
Hence option [c] is the correct answer.