# Values of Trigonometric Ratios of Typical Angles

## Values of Trigonometric Ratios of Typical Angles

1.Value of sin15°, cos15°, sin75°, cos75°, tan15°, tan75°:

(i) sin15°

sin15° = sin (45° – 30°) = sin45° cos30° – sin30° cos45°

$$\sin {{15}^{o}}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{\sqrt{2}}$$.

$$\sin {{15}^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$.

(ii) cos15°

cos15° = cos (45° – 30°) = cos45° cos30° + sin30 sin45°

$$\cos {{15}^{o}}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{\sqrt{2}}$$.

cos15° = $$\frac{\sqrt{3}+1}{2\sqrt{2}}$$.

Also, sin15° = cos75° = -cos105°

Also, cos15°= sin75° = sin105°

(iii) tan15°

Now, tan15° = tan (60° – 45°)

$$=\frac{\tan {{60}^{\circ }}-\tan {{45}^{\circ }}}{1+\tan {{60}^{\circ }}\tan {{45}^{\circ }}}$$.

$$=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$$.

And tan75° = tan (30° + 45°)

$$=\frac{\tan {{30}^{\circ }}+\tan {{45}^{\circ }}}{1-\tan {{30}^{\circ }}\tan {{45}^{\circ }}}$$.

$$=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}$$..

2. The Value of sin18°

Let θ = 18°, then 5θ = 90°

2θ + 3θ = 90°

2θ = 90° – 3θ

sin2θ = sin (90° – 3θ)

sin2θ = cos3θ

2sinθcosθ = 4cos³θ – 3cosθ

2sinθ = 4cos²θ-3 [Dividing by cosθ]

2sinθ = 4(1-sin²θ)-3 = 1-4sin²θ

4sin²θ + 2sinθ-1 = 0

$$\sin \theta =\frac{-2\pm \sqrt{4+16}}{8}=\frac{-2+2\sqrt{5}}{8}=\frac{-1\pm \sqrt{5}}{4}$$.

∵ θ = 18°

∴ sinθ = sin18° > 0

$$\therefore \,\sin \theta =\sin {{18}^{\circ }}=\frac{\sqrt{5}-1}{4}$$.

3. The Value of cos18°, cos 36°, sin 36°

(i) Value of cos18°:

$${{\cos }^{2}}{{18}^{\circ }}=1-{{\sin }^{2}}{{18}^{\circ }}=1-{{\left( \frac{\sqrt{5}-1}{4} \right)}^{2}}=1-\frac{5+1-2\sqrt{5}}{16}=\frac{10+2\sqrt{5}}{16}$$.

$$\Rightarrow \,\,\cos {{18}^{\circ }}=\frac{1}{4}\sqrt{10+2\sqrt{5}}$$ [∵ cos18° > 0].

(ii) Value of cos 36°

$$\cos {{36}^{\circ }}=1-2{{\sin }^{2}}{{18}^{\circ }}=1-2={{\left( \frac{\sqrt{5}+1}{4} \right)}^{2}}$$.

(iii) Value of sin 36°:

$${{\sin }^{2}}{{36}^{\circ }}=1-{{\cos }^{2}}{{36}^{\circ }}=1-{{\left( \frac{\sqrt{5}+1}{4} \right)}^{2}}$$.

$$=1-\frac{6+2\sqrt{5}}{16}=\frac{16-6-2\sqrt{5}}{16}=\frac{10-2\sqrt{5}}{16}$$.

$$\therefore \,\,\sin {{36}^{\circ }}=\frac{1}{4}\sqrt{10-2\sqrt{5}}$$ [∵ sin36° > 0].

Note: $$\sin {{54}^{\circ }}=\sin \left( {{90}^{\circ }}-{{36}^{\circ }} \right)=\cos {{36}^{\circ }}=\frac{\sqrt{5}+1}{4}$$.

$$\cos {{54}^{\circ }}=\cos \left( {{90}^{\circ }}-{{36}^{\circ }} \right)=\sin {{36}^{\circ }}=\frac{1}{4}\left( \sqrt{10-2\sqrt{5}} \right)$$.