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Sin Cos Formula: We all know that trigonometry is the branch of mathematics that deals with triangles. Engineering, astronomy, physics, and architectural design all benefit from trigonometric concepts. Let's look at some fundamental trigonometry sin cos formulae and trigonometric ratios in this article.
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Key Terms: Trigonometric Ratio, right-angled triangle, half angle formula , trigonometric identities, double and triple formula, Sine, Cosecant, Tangent, Cosecant, Secant, Cotangent
Basic Trigonometric Ratios Formulas
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- The right-angle triangle has six fundamental trigonometric ratios: Sine, Cosecant, Tangent, Cosecant, Secant and Cotangent.
- Sin, Cos, Tan, Cosec, Sec and Cot are the abbreviations for Sine, Cosecant, Tangent, Cosecant, Secant and Cotangent, respectively.
- The basic trigonometric ratios Sin and Cos describe the form of a right triangle.
- A right-angled triangle is one in which one of the angles is a right angle, i.e. it has a 90-degree angle.
- The hypotenuse is the side that lies opposite the right angle and it is the longest side of a right-angled triangle.
- The opposite side of the angle to be calculated is the perpendicular and the adjacent side is the base of the right-angled triangle.

Right-angle triangle
In any right-angled triangle, for any angle:
- The Sine of the Angle (sin A) = the length of the opposite side / the length of the hypotenuse = BC / AB
- The Cosine of the Angle (cos A) = the length of the adjacent side / the length of the hypotenuse = AC / AB
- The Tangent of the Angle (tan A) = the length of the opposite side /the length of the adjacent side= BC / AC
- The Cosecant of the Angle (cosec A) = the length of the hypotenuse / the length of the opposite side = AB / BC
- The Secant of the Angle (sec A) = the length of the hypotenuse / the length of the adjacent side = AB / AC
- The Cotangent of the Angle (cot A) = the length of the adjacent side / the length of the opposite side = AC / BC
Also Read:
| Related Articles | ||
|---|---|---|
| Introduction to Trigonometry | Trigonometric Identities | Trigonometry Table |
| Some Applications of Trigonometry | Trigonometric Functions | Some Applications of Trigonometry MCQs |
Reciprocal of Trigonometric Identities
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- cosec A = 1/sin A
- sec A = 1/cos A
- cot A = 1/tan A
- sin A = 1/cosec A
- cos A = 1/sec A
- tan A = 1/cot A
Read more : Cot Tan Formula
Basic Trigonometric Identities for Sin and Cos
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- cos2(A) + sin2(A) = 1
- If A + B = 180° then:
sin(A) = sin(B)
cos(A) = -cos(B)
- If A + B = 90° then:
sin(A) = cos(B)
cos(A) = sin(B)
Half-Angle Formulas for Sin and Cos
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- Sin (A/2)= ± \(\sqrt{[(1-Cos A)/2]}\)
- If A/2 is in the first or second quadrants, the value will be positive.
- If A/2 is in the third or fourth quadrants, the value will be negative.
- Cos(A/2) = ±\(\sqrt{[(1+Cos A)/2]}\)
- If A/2 is in the first or fourth quadrants, the value will be positive.
- If A/2 is in the second or third quadrants, the value will be negative.
Also Read:
Double and Triple Angle Formulas for Sin and Cos
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- Sin 2A = 2Sin A Cos A
- Cos 2A = Cos2 A – Sin2 A = 2 Cos2A - 1 = 1 - Sin2 A
- Sin 3A = 3Sin A – 4 Sin3 A
- Cos 3A = 4 Cos3 A – 3CosA
- Sin4 A = (3/8)−(1/2)cos(2A)+(1/8)cos(4A)
- Cos4 A = cos4 A – 6cos2 A sin2 A + sin4 A
- Sin2A = [1–Cos(2A)] / 2
- Cos2A = [1+Cos(2A)] / 2
Sum and Difference of Angles for Sin and Cos
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- Sin(A + B) = Sin(A).Cos(B) + Cos(A).Sin(B)
- Sin(A−B) = Sin(A)⋅Cos(B) − Cos(A)⋅Sin(B)
- Cos(A+B) = Cos(A)⋅Cos(B) − Sin(A)⋅Sin(B)
- Cos(A−B) = Cos(A)⋅Cos(B) + Sin(A)⋅Sin(B)
- Sin(A+B+C) = SinA⋅CosB⋅CosC + CosA⋅SinB⋅CosC + CosA⋅CosB⋅SinC − SinA⋅SinB⋅SinC
- Cos(A + B +C) = CosA CosB CosC - CosA SinB SinC – SinA CosB SinC – SinA SinB CosC
- Sin A + Sin B = 2 Sin[(A+B)/2] Cos[(A−B)/2]
- Sin A – Sin B = 2 Sin[(A−B)/2] Cos [(A+B)/2]
- Cos A + Cos B = 2 Cos[(A+B)/2] Cos [(A−B)/2]
- Cos A – Cos B = - 2 Sin[(A+B)/2] Sin [(A−B)/2]
Also Read: Difference between Trigonometry and Geometry
Things to Remember
- Basic Sin Cos formula is cos2(A) + sin2(A) = 1
- Sin Cos formula for half angle-
- Sin (A/2)= ±\(\sqrt{[(1-Cos A)/2]}\)
- Cos(A/2) = ±\(\sqrt{[(1+Cos A)/2]}\)
- Sin Cos formula for double angle-
- Sin 2A = 2Sin A Cos A
- Cos 2A = Cos2 A – Sin2 A = 2 Cos2A - 1 = 1 - Sin2 A
- Sin Cos formula for triple angle-
- Sin 3A = 3Sin A – 4 Sin3 A
- Cos 3A = 4 Cos3 A – 3CosA
- Sum of angles for Sin and Cos
- Sin(A + B) = Sin(A).Cos(B) + Cos(A).Sin(B)
- Cos(A+B) = Cos(A)⋅Cos(B) − Sin(A)⋅Sin(B)
- Sin(A+B+C) = SinA⋅CosB⋅CosC + CosA⋅SinB⋅CosC + CosA⋅CosB⋅SinC − SinA⋅SinB⋅SinC
- Cos(A + B +C) = CosA CosB CosC - CosA SinB SinC – SinA CosB SinC – SinA SinB CosC
- Sin A + Sin B = 2 Sin[(A+B)/2] Cos[(A−B)/2]
- Cos A + Cos B = 2 Cos[(A+B)/2] Cos [(A−B)/2]
- Differences of angles for Sin and Cos
- Sin(A−B) = Sin(A)⋅Cos(B) − Cos(A)⋅Sin(B)
- Cos(A−B) = Cos(A)⋅Cos(B) + Sin(A)⋅Sin(B)
- Sin A – Sin B = 2 Sin[(A−B)/2] Cos [(A+B)/2]
- Cos A – Cos B = - 2 Sin[(A+B)/2] Sin [(A−B)/2]
Sample Questions
Ques 1: Find the value of the trigonometric function in fraction form for triangle ABC. What is the cosine of ∠B? (1 Mark)

Ans: The cosine of an angle is the value of the adjacent side over the hypotenuse.
Therefore:
cos∠B = base/hypotenuse=7/25
Ques 2: What is the value of sin(30)+sin(60)? (2 Marks)
Ans: Solve each term separately.
Sin (30)= 1/2
Sin (60)= \(\sqrt{3}\)/2
Add both terms.
Sin (30) + Sin (60)= 1/2+\(\sqrt{3}\)/2
=(\(\sqrt{3}\)+1)/2
= 1.366
Ques 3: Determine the value of 2tan(120). (2 Marks)
Ans: Rewrite 2 tan(120) in terms of sines and cosines.
2tan(120) = 2 (sin(120) / cos(120))
= 2(\(\sqrt{3}\)/2 / −1/2)
= - 2×\(\sqrt{2}\)/2×2
=−2\(\sqrt{3}\)
Ques 4: Find the value of 1/2sin(45)+tan(60). (2 Marks)
Ans: To find the value of 1/2sin(45)+tan(60), solve each term separately.
1/2sin(45)= 1/2⋅\(\sqrt{2}\)/2 = \(\sqrt{2}\)/4
tan(60) = \(\sqrt{3}\)
Sum the two terms.
1/2sin(45)+tan(60) = \(\frac{\sqrt{2}}{4} + \sqrt{3}\)
= \(\frac{(\sqrt{2} + \frac{4}{\sqrt{3}})}{4}\)
Ques 5: In Δ ABC, right-angled at B, AB = 3 cm and AC = 6 cm. Determine angle BAC and angle ACB. (2 Marks)
Ans: Given AB = 3 cm and AC = 6 cm.
Therefore, AB/AC=sinR
or sinR=3/6=1/2
So, ∠BAC=30° and ∠ACB=60
Ques 6: What is the result when the following expression is simplified as much as possible?
(i) sin(2h)sec(h)+2sin(−h) (3 Marks)
Ans: (i) Because sinx is an odd function, we can rewrite the second term in the expression.
2sin(−h) = −2sinh.
We now use a double-angle formula to expand the first term.
sin(2h) sech = 2sinh cosh sech.
Because they are reciprocals, cosh sech = 1.
2sinh cosh sech−2sinh
= 2sinh−2sinh
= 0.
Ques 7: When, sin X = 1/2 and cos Y = 3/4 then find cos(X+Y) (3 Marks)
Ans: We know cos(X + Y) = cos X cos Y – sin X sin Y
Given sin X = 1/2
We know that, cos X = \(\sqrt{(1 - sin2X)} = \sqrt{(1 - (\frac{1}{4}))} = \frac{\sqrt{3}}{2}\)
Thus, cos X = √3/2
Given cos Y = 3/4
We know that, sin Y =\(\sqrt{(1 - cos2Y)} = \sqrt{(1 - (\frac{9}{16}))} = \frac{\sqrt{7}}{4}\)
Thus, sin Y = \(\sqrt{7}\)/4
cos X = \(\sqrt{3}\)/2, and sinY = \(\sqrt{7}\)/4
Applying the sum of cos formula, we have
cos(X+Y) = (\(\sqrt{3}\)/2) × (3/4) – 1/2 × (\(\sqrt{7}\)/4)
= \(\frac{(3\sqrt{3} -\sqrt{7})}{8}\)
Ques 8: If sin θ = 3/5, find sin2θ. (3 Marks)
Ans: We know that,
sin2θ = 2 sin θ cos θ
We need to determine cos θ.
Let us use the sin cos formula cos2θ + sin2θ = 1.
Rewriting, we get cos2θ = 1 - sin2θ
= 1-(9/25)
cos2θ = 16/25
cos θ = 4/5
sin2θ = 2sinθcos θ
= 2 × (3/5) × (4/5) = 24/25
Ques 9: Prove that: \(\frac{tan A + sec A -1}{tan A -secA +1}\) =\(\frac{1 +sinA}{cosA}\)
Ans:

Ques 10: Find the value of the expression cos4π/8 + cos43π/8 + cos45π/8 + cos47π/8 (5 Marks)
Ans:

Ques 11: If \(tan \theta = \frac{sin \alpha - cos \alpha}{sin \alpha + cos \alpha}\), then show sin α + cos α =\(\sqrt{2} cos \theta\)
Ans:

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