Sin Cos Formulas: Basic Trigonometric Ratios and Identities

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Jasmine Grover

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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.

Check out: Who is the father of mathematics?

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. 

The right-angle 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:


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)
triangle ABC

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: 

tan A + sec A- 1tan A - sec A +1 = 1+ sin Acos A

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

Ans: 

Find the value of the expression cos4 ?/8 + cos4 3?/8  + cos4 5?/8  + cos47?/8

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: 

If tan? = sin ? - cos ? sin ? + cos ? , then show sin ? + cos ? =2 cos?

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CBSE CLASS XII Related Questions

  • 1.

    At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


    Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
    On the basis of the above information, answer the following questions :


      • 2.
        Find:

        If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

          • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
          • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
          • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
          • \(p = 0, \, q = 0\)

        • 3.

          A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length. 


            • 4.
              Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


                • 5.
                  Find:

                  The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]


                    • 6.
                      Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).

                        CBSE CLASS XII Previous Year Papers

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