## Trigonometric Ratios of Submultiple Angles:

The angles (1/2) A, (1/3) A, ……etc., are called the submultiple angles of angle A. For all values of the angle A, we have-

sin 2A = 2 sin A cos A = 2 tan A/(1 + tan^{2} A)cos 2A = cos ^{2} A – sin^{2} A = 2 cos^{2} A – 1 = 1 – 2 sin^{2} A = (1 – tan^{2} A)/(1 + tan^{2} A)tan 2A = 2 tan A/(1 – tan ^{2} A)cot 2A = (cot ^{2} A – 1)/2 cot A |

Replacing 2A by θ in each of the above formulae, we get-

sin θ = 2 sin θ/2 cos θ/2 = (2 tan θ/2)/(1 + tan^{2} θ/2) ……………(ii)cos θ = cos ^{2} θ/2 – sin^{2} θ/2 = 2 cos^{2} θ/2 – 1 = 1 – 2 sin^{2} θ/2 = (1 – tan^{2} θ/2)/(1 + tan^{2} θ/2) ………(iii)tan θ = (2 tan θ/2)/(1 – tan ^{2} θ/2) …………(iv)cot θ = (cot ^{2} θ/2 – 1)/(2 cot θ/2) ………….(v) |

Again, for all values of the angle A, we can write-

sin 3A = 3 sin A – 4 sin^{3} Acos 3A = 4 cos ^{3} A – 3 cos Atan 3A = (3 tan A -tan ^{3} A)/(1 – 3 tan^{2} A) |

Replacing 3A by θ in each of the above formulae, we get-

sin θ = 3 sin θ/3 – 4 sin^{3} θ/3 ………….(vii)cos θ = 4 cos ^{3} θ/3 – 3 cos θ/3 ………….(viii)tan θ = (3 tan θ/3 -tan ^{3} θ/3)/(1 – 3 tan^{2} θ/3) …………..(ix) |

If θ is known, we can find out the respective quadrants in which (π/4 + θ) lies. Then, using the convention ‘all, sin, tan, cos’, we can determine the appropriate signs on the right side of equations (xvii) and (xviii).

If the value of θ is known, we can find the quadrant in which θ/2 lies. Then, using the convention ‘all, sin,

tan, cos’, we can determine the appropriate sign of tan θ/2.

## To find the Trigonometrical Ratios of 18°, 36°, 54° and 72°:

## To find the Trigonometrical Ratios of 3°, 6°, 9° and 12°:

Using a similar method, we can find sines and cosines of 6°, 9°, 12° etc. Clearly, 6° = 36° – 30°, 9° = 45° – 36°, 12° = 30° – 18°, 21° = 36° – 15° and so on.

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