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Trigonometric Ratios

If θ = 30o,ve...

Question

If $θ = 30^{o}$ ,verify that:

(i) $t a n 2 θ = \frac{2 t a n θ}{1 - t a n^{2} θ}$

(ii) $s i n 2 θ = \frac{2 t a n θ}{1 + t a n^{2} θ}$

(iii) $c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}$

(iv) $c o s 3 θ = 4 c o s^{3} θ - 3 c o s θ$

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Solution

$left parenthesis i right parenthesis space L H S thin space equals space tan 2 theta equals space tan 60 space equals space square root of 3 R H S space equals space fraction numerator 2 tan theta over denominator 1 minus tan squared theta end fraction space equals fraction numerator 2 tan 30 over denominator 1 minus tan squared 30 end fraction space equals space fraction numerator 2 begin display style fraction numerator 1 over denominator square root of 3 end fraction end style over denominator 1 minus begin display style 1 third end style end fraction space equals space fraction numerator begin display style fraction numerator 2 over denominator square root of 3 end fraction end style over denominator begin display style 2 over 3 end style end fraction space equals space square root of 3 h e n c e space L H S thin space equals space R H S space h e n c e space v e r i f i e d$
$left parenthesis i i right parenthesis L H S space equals space sin 2 theta space equals sin 60 space equals space fraction numerator square root of 3 over denominator 2 end fraction R H S space equals space fraction numerator 2 tan theta over denominator 1 plus tan squared theta end fraction space equals space fraction numerator 2 tan 30 over denominator 1 plus tan squared 30 end fraction space equals space fraction numerator 2 begin display style fraction numerator 1 over denominator square root of 3 end fraction end style over denominator 1 plus open parentheses begin display style fraction numerator 1 over denominator square root of 3 end fraction end style close parentheses squared end fraction space equals fraction numerator begin display style fraction numerator 2 over denominator square root of 3 end fraction end style over denominator 1 plus begin display style 1 third end style end fraction space equals space fraction numerator begin display style fraction numerator 2 over denominator square root of 3 end fraction end style over denominator begin display style 4 over 3 end style end fraction space equals space fraction numerator square root of 3 over denominator 2 end fraction L H S space equals space R H S space h e n c e space v e r i f i e d$

$left parenthesis i i i right parenthesis space L H S space equals space cos 2 theta space equals cos 60 space equals space 1 half R H S space equals space fraction numerator 1 minus tan squared theta over denominator 1 plus tan squared theta end fraction space equals space fraction numerator 1 minus tan squared 30 over denominator 1 plus tan squared 30 end fraction space equals space fraction numerator 1 minus begin display style 1 third end style over denominator 1 plus begin display style 1 third end style end fraction space equals space fraction numerator begin display style 2 over 3 end style over denominator begin display style 4 over 3 end style end fraction space equals space 1 half L H S space equals space R H S space h e n c e space v e r i f i e d.$

$left parenthesis i v right parenthesis space L H S space equals space cos 3 theta space equals space cos 90 space equals space 0 R H S space equals space 4 cos cubed theta minus 3 cos theta space equals space 4 cos cubed 30 space minus space 3 cos 30 equals space 4 open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses cubed minus 3 fraction numerator square root of 3 over denominator 2 end fraction space equals space fraction numerator 4 cross times 3 square root of 3 over denominator 8 end fraction minus 3 fraction numerator square root of 3 over denominator 2 end fraction space equals 0 L H S thin space equals space R H S space h e n c e space v e r i f i e d.$

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Similar questions

θ = 30^{\circ},

verify the following:

i) cos 3 θ = 4 {cos}^{3} θ - 3 cos θ

i i) sin 3 θ = 3 sin θ - 4 {sin}^{3} θ

\frac{{cos}^{2} θ + {tan}^{2} θ - 1}{{sin}^{2} θ} = {tan}^{2} θ

θ

is an acute angle and

sin θ = cos θ

3 {tan}^{2} θ + 2 {sin}^{2} θ + {cos}^{2} θ - 1

sin 2 θ

\frac{2 tan θ}{1 - {tan}^{2} θ}

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

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