Trigonometry is the study of relationships that deal with angles, lengths, and heights of triangles and relations between different parts of circles and other geometrical figures. Concepts of trigonometry are very useful in engineering, astronomy, Physics, and architectural design. Sine and cos are the terms used in the trigonometry that tell us about the triangle. Let us discuss in detail about the sin cos formula and other concepts.

**Sin Cos Formula**

**Basic trigonometric ratios**

There are six trigonometric ratios for the right angle triangleÂ are Sin, Cos, Tan, Cosec, Sec, Cot which stands for Sine, Cosecant, Tangent, Cosecant, Secant respectively.Â Sin and Cos are basic trigonometric functions that tell about the shape of a right triangle. SO let us see the sin cos formula along with the other important trigonometric ratios.

- \(\sin Î¸= Perpendicular/ Hypotenuse\)
- \(\cos Î¸= Base/ Hypotenuse\)
- \(\tan Î¸= Perpendicular/Base\)
- \(\csc Î¸= Hypotenuse/Perpendicular\)
- \(\sec Î¸ = Hypotenuse/Base\)
- \(\cot Î¸= Base/Perpendicular\)

**Basic Trigonometric Identities for Sine and Cos**

- \(\cos^2 (A) + \sin^2 (A) = 1\)

If A + B = 180Â° then:

- \(\sin(A) = \sin(B)\)
- \(\cos(A) = -\cos(B)\)

If A + B = 90Â° then:

- \(\sin(A) = \cos(B)\)
- \(\cos(A) = \sin(B)\)

**Half-angle formulas**

\(\sin(\frac{A}{2}) =Â \sqrt{\frac{Â±1âˆ’\cos(A) }{2}}\)

- If \(\frac{A}{2}\) lies in quadrant I or II
- If \(\frac{A}{2}\) lies in quadrant III or IV

\(\cos (\frac{A}{2}) =Â \sqrt{ \frac{Â±1+\cos(A)}{2}}\)

- If \(\frac{A}{2}\) lies in quadrant I or IV
- If \(\frac{A}{2}\) lies in quadrant II or III

**Double and Triple Angle Formulas**

- \(\sin 2A = 2\sin A \cos A\)
- \(\cos 2A = \cos^2 A â€“ \sin^2 A = 2 \cos^2Â â€“ 1 = 1- \sin^2 A\)
- \(\sin 3A = 3\sin A â€“ 4 \sin ^3 A\)
- \(\cos 3A = 4 \cos^3 A â€“ 3\cos A\)
- \(\sin^4 A = 4 \cos^3 AÂ \sin A â€“ 4\cos A \sin^3 A\)
- \(\cos^4Â A = \cos^4 A â€“ 6\cos^2 A\sin^2 A +\sin^4 A\)
- \(\sin^2 A =Â \frac{1â€“\cos(2A)}{2}\)
- \(\cos^2 A = \frac{1+\cos(2A)}{2}\)

**Sum and Difference of Angles**

- \(\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)\)
- \(\sin(B)\sin(Aâˆ’B)=\sin(A)\cos(B)âˆ’\cos(A)\sin(B)\)
- \(\cos(A+B)=\cos(A)\cos(B)âˆ’\sin(A)\sin(B)\)
- \(\cos(Aâˆ’B)=\cos(A)\cos(B)+\sin(A)\sin(B)\)
- \(\sin(A+B+C)=\sin A\cos B\cos C+\cos A\sin B\cos C+\cos A\cos B\sin Câˆ’\sin A\sin B\sin C\)
- \(\cos(A+B+C)=\cos A\cos B\cos Câˆ’\sin A\sin B\cos Câˆ’\sin A\cos B\sin Câˆ’\sin A\cos B\sin Câˆ’\cos A\sin B\sin C\)
- \(\sin A + \sin B = 2\sin \frac{(A+B)}{2}\cos \frac{(Aâˆ’B)}{2}\)
- \(\sin A â€“ \sin B = 2\sin \frac{(Aâˆ’B)}{2}\cos \frac{(A+B)}{2}\)
- \(\cos A + \cos B = 2\cos\frac{(A+B)}{2} (A+B)2\cos\frac{(Aâˆ’B)}{2}\)
- \(\cos A + \cos B = -2\sin \frac{(A+B)}{2}\sin \frac{(Aâˆ’B)}{2}\)

**Product Identities**

- \(\sin(x) \cos(y) = \frac{1}{2} [\sin(x + y) + \sin(x – y)]\)
- \(\cos(x) \sin(y) = \frac{1}{2} [\sin(x + y) – \sin(x – y)]\)
- \(\cos(x) \cos(y) = \frac{1}{2} [\cos(x – y) + \cos(x + y)]\)
- \(\sin(x) \sin(y) = \frac{1}{2} [\cos(x – y) – \cos(x + y)]\)

## Solved Examples

Q.1. In a right triangle ABC, \(\tan A = 3/4. Find \sin A and \cos A.\)

Solution: Let a be the length of the side opposite angle A, b the length of the side adjacent to angle A and h be the length of the hypotenuse.

\(\tan A = \frac{ opposite side }{ adjacent side } Â = \frac{a}{b} = \frac{3}{4}\)

we can say that: a = 3k and b = 4k , where k is a coefficient of proportionality.

Pythagorasâ€™s theorem:

h^{2}Â = (3k)^{2}Â + (4k)^{2}

h = 5k

\(\sin A = \frac{a}{h} = \frac{3k}{5k} = \frac{3}{5}\) and

\(\cos A = \frac{4k}{5k}= \frac{4}{5}\)

Q2. In Î” ABC, right-angled at B, AB = 3 cm and AC = 6 cm. Determine \angle BAC and \angle ACB.

Solution: Given AB = 3 cm and AC = 6 cm.

Therefore, \(\frac{AB}{AC}=\sin R\)

or \(\sin R = \frac{3}{6}= \frac{1}{2}\)

So, \(\angle BAC = 30Â° andÂ \angle ACB = 60\)

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26