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Question

Question: Find the value of \(\sin \left( 270{}^\circ \right)\)....

Find the value of sin(270)\sin \left( 270{}^\circ \right).

Explanation

Solution

Hint: The value of sinθ\sin \theta is negative in the third quadrant. Also, we can write sin(270)\sin \left( 270{}^\circ \right) in the form of sin(180+θ)\sin \left( 180{}^\circ +\theta \right). We will then apply the formula of sin(π+θ)=sinθ\sin \left( \pi +\theta \right)=-\sin \theta to find the value of sin(270)\sin \left( 270{}^\circ \right).

Complete step-by-step answer:
It is given in the question that we have to find the value of sin(270)\sin \left( 270{}^\circ \right). We know that the value of sinθ\sin \theta increases with the increase in the value of θ\theta . We also know that the value of sinθ\sin \theta is negative in the third quadrant. We also know that, sin(π+θ)=sinθ\sin \left( \pi +\theta \right)=-\sin \theta . So, we will use this formula to find the value of sin(270)\sin \left( 270{}^\circ \right).

Now, we know that we can also write sin(270)\sin \left( 270{}^\circ \right) as sin(180+90)\sin \left( 180{}^\circ +90{}^\circ \right). So, applying the formula, we will get,
sin(180+90)=sin90\sin \left( 180{}^\circ +90{}^\circ \right)=-\sin 90{}^\circ
We know that the value of sin90=1\sin 90{}^\circ =1, so sin90=1-\sin 90{}^\circ =-1. So, we by applying that, we will get,
sin(180+90)=1\sin \left( 180{}^\circ +90{}^\circ \right)=-1 or sin(270)=1\sin \left( 270{}^\circ \right)=-1.
Therefore, we get the value of sin(270)=1\sin \left( 270{}^\circ \right)=-1.

Note: The students must remember the trigonometric formulas to answer such questions. They should not get confused with the formulas and miss out on any signs, like they should not write the formula as, sin(π+θ)=sinθ\sin \left( \pi +\theta \right)=\sin \theta as that would lead to an incorrect answer. We can also solve this using an alternate method. We can use the formula, sin(2πθ)=sinθ\sin \left( 2\pi -\theta \right)=-\sin \theta . So, we can express sin(270)\sin \left( 270{}^\circ \right) as sin(36090)=sin90=1\sin \left( 360{}^\circ -90{}^\circ \right)=-\sin 90{}^\circ =-1.