Questions & Answers

Question

Answers

(a) 1

(b) 0

(c) -1

(d) $\dfrac{1}{2}$

Answer
Verified

Hint: Here we have to use the values of trigonometric ratios for a particular angle. And we have to use one property that is multiplication of anything with 0 is always 0.

Complete step-by-step answer:

As you can see in \[\cos {1^ \circ } \cdot \cos {2^ \circ } \cdot \cos {3^ \circ }.......\cos {180^ \circ }\]

It is product of cosine of all angle from ${1^ \circ }$ to ${180^ \circ }$

In which ${30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$ etc, most angles come as you know the value of the cosine of these angles.

AS you know the value of

$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\cos {60^ \circ } = \dfrac{1}{2},\cos {90^ \circ } = 0$

We can write question like this

\[ \Rightarrow \cos {1^ \circ } \cdot \cos {2^ \circ } \cdot \cos {3^ \circ }......\cos {30^ \circ }....\cos {45^ \circ }....\cos {60^ \circ }.....\cos {90^ \circ }.....\cos {180^ \circ }\]

All values are in multiply so you know value of $\cos {90^ \circ } = 0$

\[ \Rightarrow \cos {1^ \circ } \cdot \cos {2^ \circ } \cdot \cos {3^ \circ }......\cos {30^ \circ }....\cos {45^ \circ }....\cos {60^ \circ }..... \times 0 \times .....\cos {180^ \circ }\]

As you know the multiple of 0 from any number then result comes to 0

$ \Rightarrow 0$

Answer is 0.

So, the correct option is (b).

Note: Whenever you come to these type of problem you have to use known value of trigonometric angle (like \[\sin {30^ \circ },\tan {45^ \circ }\] etc) and try to use these values in equation after some rearrangement then you can easily get answer.

Complete step-by-step answer:

As you can see in \[\cos {1^ \circ } \cdot \cos {2^ \circ } \cdot \cos {3^ \circ }.......\cos {180^ \circ }\]

It is product of cosine of all angle from ${1^ \circ }$ to ${180^ \circ }$

In which ${30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$ etc, most angles come as you know the value of the cosine of these angles.

AS you know the value of

$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\cos {60^ \circ } = \dfrac{1}{2},\cos {90^ \circ } = 0$

We can write question like this

\[ \Rightarrow \cos {1^ \circ } \cdot \cos {2^ \circ } \cdot \cos {3^ \circ }......\cos {30^ \circ }....\cos {45^ \circ }....\cos {60^ \circ }.....\cos {90^ \circ }.....\cos {180^ \circ }\]

All values are in multiply so you know value of $\cos {90^ \circ } = 0$

\[ \Rightarrow \cos {1^ \circ } \cdot \cos {2^ \circ } \cdot \cos {3^ \circ }......\cos {30^ \circ }....\cos {45^ \circ }....\cos {60^ \circ }..... \times 0 \times .....\cos {180^ \circ }\]

As you know the multiple of 0 from any number then result comes to 0

$ \Rightarrow 0$

Answer is 0.

So, the correct option is (b).

Note: Whenever you come to these type of problem you have to use known value of trigonometric angle (like \[\sin {30^ \circ },\tan {45^ \circ }\] etc) and try to use these values in equation after some rearrangement then you can easily get answer.

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