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Question: Evaluate the following \(\left( \cos 0{}^\circ +\sin 45{}^\circ +\sin 30{}^\circ \right)\left( \sin ...

Evaluate the following (cos⁡0∘+sin⁡45∘+sin⁡30∘)(sin⁡90∘−cos⁡45∘+cos⁡60∘)\left( \cos 0{}^\circ +\sin 45{}^\circ +\sin 30{}^\circ \right)\left( \sin 90{}^\circ -\cos 45{}^\circ +\cos 60{}^\circ \right).
A. 54\dfrac{5}{4}
B. 94\dfrac{9}{4}
C. 74\dfrac{7}{4}
D. None of these

Explanation

Solution

Hint: To solve this question, we should know few of the trigonometric values or we can say the trigonometric ratios like sin⁡0∘=cos⁡90∘=0,sin⁡30∘=cos⁡60∘=12,sin⁡45∘=cos⁡45∘=12,sin⁡60∘=cos⁡30∘=32\sin 0{}^\circ =\cos 90{}^\circ =0,\sin 30{}^\circ =\cos 60{}^\circ =\dfrac{1}{2},\sin 45{}^\circ =\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\sin 60{}^\circ =\cos 30{}^\circ =\dfrac{\sqrt{3}}{2} and sin⁡90∘=cos⁡0∘=1\sin 90{}^\circ =\cos 0{}^\circ =1. By using these values, we can find the answer to the given question.

Complete step-by-step answer:

In this question, we have been asked to find the value of(cos⁡0∘+sin⁡45∘+sin⁡30∘)(sin⁡90∘−cos⁡45∘+cos⁡60∘)\left( \cos 0{}^\circ +\sin 45{}^\circ +\sin 30{}^\circ \right)\left( \sin 90{}^\circ -\cos 45{}^\circ +\cos 60{}^\circ \right). We know that cos⁡0∘=0,sin⁡45∘=12,sin⁡30∘=12,sin⁡90∘=1,cos⁡45∘=12\cos 0{}^\circ =0,\sin 45{}^\circ =\dfrac{1}{\sqrt{2}},\sin 30{}^\circ =\dfrac{1}{2},\sin 90{}^\circ =1,\cos 45{}^\circ =\dfrac{1}{\sqrt{2}} and cos⁡60∘=12\cos 60{}^\circ =\dfrac{1}{2}. So, we will use these values to find the value of the expression that is given in the question. So, by substituting these values in the expression given in our question, we get,
[1+12+12][1−12+12]\left[ 1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right]\left[ 1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right]
Now, we will find the LCM. By taking the LCM in both factors of the above expression, we get,
[22+2+222][22−2+222]\left[ \dfrac{2\sqrt{2}+2+\sqrt{2}}{2\sqrt{2}} \right]\left[ \dfrac{2\sqrt{2}-2+\sqrt{2}}{2\sqrt{2}} \right]
On further simplification of the above expression, we get,
[32+222][32−222]\left[ \dfrac{3\sqrt{2}+2}{2\sqrt{2}} \right]\left[ \dfrac{3\sqrt{2}-2}{2\sqrt{2}} \right]
Now, we know that (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}. So, by applying that in the above equation, we get,
[(32)2−(2)2(22)2]\left[ \dfrac{{{\left( 3\sqrt{2} \right)}^{2}}-{{\left( 2 \right)}^{2}}}{{{\left( 2\sqrt{2} \right)}^{2}}} \right]
On further simplification of the above expression, we get,
[18−48]=148=74\left[ \dfrac{18-4}{8} \right]=\dfrac{14}{8}=\dfrac{7}{4}
Therefore, we have obtained the value of the given expression as 74\dfrac{7}{4}.
Hence, the correct answer for the expression given in the question is option C.

Note: The possible mistakes that the students can make while solving this question are the simple calculation mistakes. They may also not take the least common multiple or the LCM carefully which may lead to the wrong answer. We can also solve this question by writing the terms in second factor as cos⁡(90∘−θ)\cos \left( 90{}^\circ -\theta \right) if it is in sin⁡θ\sin \theta and as sin⁡(90∘−θ)\sin \left( 90{}^\circ -\theta \right) if it is in cos⁡θ\cos \theta . We can then apply (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} to get the desired value.