Trigonometric Identities

  • {\sin ^2}\theta  + {\cos ^2}\theta  = 1
  • 1 + {\cot ^2}\theta  = \cos e{c^2}\theta
  • 1 + {\tan ^2}\theta  = {\sec ^2}\theta

Example: In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm.

Determine:
(i) sin A, cos A
(ii) sin C, cos C

Solution:

In a given triangle ABC, right-angled at B =\angle B = 90^\circ

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

A{C^2} = A{B^2} + B{C^2}

A{C^2} = {(24)^2} + {7^2}

A{C^2} = (576 + 49)

A{C^2} = {625^{}}c{m^2}

Therefore, AC = 25 cm

(i) Sin A = BC/AC = 7/25

cos A = AB/AC = 24/25

(ii).Sin C = AB/AC = 24/25

Cos C = BC/AC = 7/25

Example If Sin A = 3/4, Calculate cos A and tan A.

Sin A = 3/4

Sin A = Opposite Side/Hypotenuse Side = 3/4

Now, let BC be 3k and AC will be 4k.

As per the Pythagoras theorem, we know;

Hypotenuse^2 = Perpendicular^2+ Base^2

A{C^2} = A{B^2} + B{C^2}

Substitute the value of AC and BC in the above expression to get;

{(4k)^2} = {(AB)^2} + {(3k)^2}

16{k^2} - 9{k^2} = A{B^2}

A{B^2} = 7{k^2}

Hence, AB = \sqrt 7 k

cos A = Adjacent Side/Hypotenuse side = AB/AC

cos A = \sqrt 7 k/4k = \sqrt 7 /4

And,

Tan A = Opposite side/Adjacent side = BC/AB

Tan A = 3k/\sqrt 7 k = 3/\sqrt 7

Example: If \angle A and \angle B are acute angles such that cos A = cos B, then show that \angle A = \angle B.

cos A = AC/AB

cos B = BC/AB

Since, it is given,

cos A = cos B

AC/AB = BC/AB

AC = BC

We know that by isosceles triangle theorem, the angles opposite to the equal sides are equal.

Therefore, \angle A = \angle B

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