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Trigonometric Functions

Short Answer Type

601. Deduce the sine formula from cosine formula.

93 Views

Long Answer Type

602. In any triangle ABC, prove that

straight a over sinA space equals space straight b over sinB space equals space fraction numerator straight c over denominator sin space straight C end fraction comma

i.e. sides of the triangle are proportional to the sines of the opposite angles.

103 Views

603. In any triangle ABC, prove that

cos space straight A space equals space fraction numerator straight b squared plus straight c squared minus straight a squared over denominator 2 bc end fraction comma space cos space straight B space equals space fraction numerator straight c squared plus straight a squared minus straight b squared over denominator 2 ac end fraction space and space cos space straight C space equals space fraction numerator straight a squared plus straight b squared minus straight c squared over denominator 2 ab end fraction

where a, b and c are as usual the sides of the triangle.

108 Views

604.

In any triangle ABC, prove that:

(i) a = b cos C + c cos B (ii) b = c cos A + a cos C (iii) c = a cos B + b cos A

118 Views

Short Answer Type

605.

Prove that in any triangle ABC.

$space space straight capital delta space equals space 1 half bc space sinA space equals space 1 half ac space sinB space equals space 1 half ab space sinC equals square root of straight s left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis left parenthesis straight s minus straight c right parenthesis end root$

98 Views

606.
In any triangle ABC, prove that

$space space space straight capital delta space equals space 1 half straight a squared fraction numerator sinB space sinC over denominator sinA end fraction equals 1 half straight b squared fraction numerator sinC space sinA over denominator sin space straight B end fraction space equals space 1 half straight c squared fraction numerator sinA space sinB over denominator sin space straight C end fraction$

Proof:

We have, $straight capital delta space equals space 1 half bc space sinA$

$space space equals 1 half fraction numerator straight k space sinB space straight k space sinC over denominator sinA end fraction sin squared straight A$ [By sine formula]

$equals space 1 half left parenthesis straight k space sinA right parenthesis squared fraction numerator sinB space sinC over denominator sin space straight A end fraction space equals space 1 half straight a squared open parentheses fraction numerator sinB space sinC over denominator sinA end fraction close parentheses$

Similarly, we can show that

$straight capital delta space equals space 1 half straight b squared fraction numerator sinC space sinA over denominator sin space straight B end fraction equals 1 half straight c squared fraction numerator sinA space sinB over denominator sin space straight C end fraction$

101 Views

607.

Prove that:

$straight s left parenthesis straight s minus straight a right parenthesis space tan straight A over 2 equals straight capital delta$

90 Views

Long Answer Type

608.

Prove that for any triangle ABC

$2 straight R space equals space fraction numerator straight a over denominator sin space straight A end fraction equals straight b over sinB equals fraction numerator straight c over denominator sin space straight C end fraction$

where R is the radius of the circumcircle.

89 Views

Short Answer Type

609.

Prove that in any triangle ABC

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