Question

Find the value of \[\cos {{300}^{\circ }}-\sin {{420}^{\circ }}+\dfrac{\tan {{120}^{\circ }}}{2}+\cos \left( -{{30}^{\circ }} \right)\].
A. \[\dfrac{1+\sqrt{3}}{2}\].
B. \[\dfrac{\sqrt{3}-1}{2}\].
C. \[\dfrac{1-\sqrt{3}}{2}\].
D. \[\dfrac{2+\sqrt{3}}{2}\].

Answer 1

Hint: Here we have to find the value of the given trigonometric degree values. We can first split the degree values for the first two terms, we can then find the degree values for every term from the trigonometric table values and substitute it. We can then simplify it step by step to get the value of the given trigonometric expression.

Complete step by step solution:
Here we have to find the value for,
\[\cos {{300}^{\circ }}-\sin {{420}^{\circ }}+\dfrac{\tan {{120}^{\circ }}}{2}+\cos \left( -{{30}^{\circ }} \right)\]
We can now split the degree values for the first two term, we get the form’
\[\Rightarrow \cos \left( {{360}^{\circ }}-{{60}^{\circ }} \right)-\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right)+\dfrac{\tan {{120}^{\circ }}}{2}+\cos \left( -{{30}^{\circ }} \right)\]
We know that the first two terms can be written as,
\[\Rightarrow \cos \left( {{60}^{\circ }} \right)-\sin \left( {{60}^{\circ }} \right)+\dfrac{\tan {{120}^{\circ }}}{2}+\cos \left( -{{30}^{\circ }} \right)\]
We can now write the degree values from the trigonometric degree values table.
We know that \[\tan {{120}^{\circ }}=-\sqrt{3},\cos \left( -{{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}\] (since \[\cos \left( -\theta \right)=\cos \theta \]) , we can now substitute these values in the above step, we get
 \[\Rightarrow \cos \left( {{60}^{\circ }} \right)-\sin \left( {{60}^{\circ }} \right)-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\]
We can now cancel the similar terms in the above step, we get
\[\Rightarrow \cos \left( {{60}^{\circ }} \right)-\sin \left( {{60}^{\circ }} \right)\]
We know that the value of \[\cos {{60}^{\circ }}=\dfrac{1}{2}\] and \[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\].
We can now substitute these values in the above step, we get
 \[\Rightarrow \dfrac{1}{2}-\dfrac{\sqrt{3}}{2}=\dfrac{1-\sqrt{3}}{2}\]

So, the correct answer is “Option C”.

Note: We should always remember some of the degree values of sine, cosine and tangent to solve these types of problems. We should also know to convert the larger degree values to its equivalent smaller values by splitting into two and solving it. We should remember that \[\cos \left( -\theta \right)=\cos \theta \].