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Trigonometric Functions

Short Answer Type

601. Deduce the sine formula from cosine formula.

We have,

$fraction numerator straight a squared over denominator sin squared straight A end fraction equals fraction numerator straight a squared over denominator 1 minus cos squared straight A end fraction equals fraction numerator straight a squared over denominator 1 minus open square brackets begin display style fraction numerator straight b squared plus straight c squared minus straight a squared over denominator 2 bc end fraction end style close square brackets squared end fraction equals fraction numerator straight a squared over denominator begin display style fraction numerator 4 straight b squared straight c squared minus left parenthesis straight b squared plus straight c squared minus straight a squared right parenthesis squared over denominator 4 straight b squared straight c squared end fraction end style end fraction$

 $equals fraction numerator 4 straight a squared straight b squared straight c squared over denominator left parenthesis 2 bc plus straight b squared plus straight c squared minus straight a squared right parenthesis space left parenthesis 2 bc minus straight b squared minus straight c squared plus straight a squared right parenthesis end fraction$

 $equals fraction numerator 4 straight a squared straight b squared straight c squared over denominator left square bracket left parenthesis straight b plus straight c right parenthesis squared minus straight a squared right square bracket space left square bracket straight a squared minus left parenthesis straight b minus straight c right parenthesis squared right square bracket end fraction$

$equals space fraction numerator 4 straight a squared straight b squared straight c squared over denominator left parenthesis straight b plus straight c plus straight a right parenthesis left parenthesis straight b plus straight c minus straight a right parenthesis left parenthesis straight a plus straight b minus straight c right parenthesis left parenthesis straight a minus straight b plus straight c right parenthesis end fraction$

$fraction numerator straight a over denominator sin space straight A end fraction space equals space fraction numerator 2 abc over denominator square root of 2 straight s left parenthesis 2 straight s minus 2 straight a right parenthesis space left parenthesis 2 straight s minus 2 straight c right parenthesis space left parenthesis 2 straight s minus 2 straight b right parenthesis end root end fraction$

 $equals space fraction numerator abc over denominator 2 square root of straight s left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis left parenthesis straight s minus straight c right parenthesis end root end fraction$ [∵ sin A is positive as $straight A less than 180 degree$ ]

Similarly, we can show

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Hence, $straight a over sinA equals straight b over sinB equals straight c over sinC$

Note: Area of the traingle ABC is denoted by A.

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

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

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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119 Views

610. Prove that in any triangle ABC

space space space space space space straight r space equals space straight capital delta over straight s

114 Views