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Notely Maths 12
Class 12 Maths
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7.3 Q-11

Question Statement

Prove that the trigonometric identity for the cosine of the sum of two angles holds:

cos⁡(α+β)=cos⁡α⋅cos⁡β−sin⁡α⋅sin⁡β\cos(\alpha + \beta) = \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta

Background and Explanation

This problem involves proving a well-known identity in trigonometry: the cosine of a sum. To prove this, we’ll use the geometric interpretation of trigonometric functions and the properties of the dot product of vectors. Specifically, we will express the vectors corresponding to the angles α\alpha and β\beta and use their dot product to arrive at the identity.

Solution

Step 1: Define unit vectors along the angles α\alpha and β\beta

Let a^\hat{a} and b^\hat{b} be unit vectors making angles α\alpha and β\beta with the xx-axis, respectively. The position vectors of these unit vectors in terms of the xx- and yy-coordinates are:

  1. For a^\hat{a} making an angle α\alpha with the xx-axis:
a^=cos⁡α,i^+sin⁡α,j^\hat{a} = \cos \alpha , \hat{i} + \sin \alpha , \hat{j}

This represents a unit vector in the direction of angle α\alpha, where i^\hat{i} and j^\hat{j} are the unit vectors along the xx- and yy-axes, respectively.

  1. For b^\hat{b} making an angle β\beta with the xx-axis:
b^=cos⁡β,i^+sin⁡β,j^\hat{b} = \cos \beta , \hat{i} + \sin \beta , \hat{j}

This represents a unit vector in the direction of angle β\beta.

Step 2: Use the dot product to relate the angles

The cosine of the angle between the two vectors a^\hat{a} and b^\hat{b} is given by the dot product of the two unit vectors:

a^⋅b^=(cos⁡α,i^+sin⁡α,j^)⋅(cos⁡β,i^+sin⁡β,j^)\hat{a} \cdot \hat{b} = (\cos \alpha , \hat{i} + \sin \alpha , \hat{j}) \cdot (\cos \beta , \hat{i} + \sin \beta , \hat{j})

Step 3: Simplify the dot product

Using the distributive property of the dot product:

a^⋅b^=cos⁡α⋅cos⁡β+sin⁡α⋅sin⁡β\hat{a} \cdot \hat{b} = \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta

This is the standard expression for the dot product of two vectors, where:

  • i^⋅i^=1\hat{i} \cdot \hat{i} = 1
  • j^⋅j^=1\hat{j} \cdot \hat{j} = 1
  • i^⋅j^=0\hat{i} \cdot \hat{j} = 0 (since they are perpendicular)

Thus:

a^⋅b^=cos⁡α⋅cos⁡β+sin⁡α⋅sin⁡β\hat{a} \cdot \hat{b} = \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta

Step 4: Conclusion

The angle between the two vectors a^\hat{a} and b^\hat{b} is α+β\alpha + \beta, so by the definition of the cosine of the angle between two vectors:

cos⁡(α+β)=a^⋅b^\cos(\alpha + \beta) = \hat{a} \cdot \hat{b}

Substitute the expression for a^⋅b^\hat{a} \cdot \hat{b}:

cos⁡(α+β)=cos⁡α⋅cos⁡β−sin⁡α⋅sin⁡β\cos(\alpha + \beta) = \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta

Thus, we have proven the trigonometric identity:

cos⁡(α+β)=cos⁡α⋅cos⁡β−sin⁡α⋅sin⁡β\cos(\alpha + \beta) = \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta

Key Formulas or Methods Used

  • Dot product of vectors: The dot product of two vectors u=(u1,u2)\mathbf{u} = (u_1, u_2) and v=(v1,v2)\mathbf{v} = (v_1, v_2) is:
u⋅v=u1v1+u2v2\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2

For unit vectors a^\hat{a} and b^\hat{b}, the dot product can be written as:

a^⋅b^=cos⁡(θ)\hat{a} \cdot \hat{b} = \cos(\theta)

where θ\theta is the angle between them.

  • Vector components: The components of a vector in terms of its direction cosine and sine are given by:
a^=cos⁡α,i^+sin⁡α,j^,b^=cos⁡β,i^+sin⁡β,j^\hat{a} = \cos \alpha , \hat{i} + \sin \alpha , \hat{j}, \quad \hat{b} = \cos \beta , \hat{i} + \sin \beta , \hat{j}

Summary of Steps

  1. Define the unit vectors a^\hat{a} and b^\hat{b} in terms of angles α\alpha and β\beta.
  2. Calculate the dot product of the two unit vectors.
  3. Simplify the dot product to find the relation between cos⁡(α+β)\cos(\alpha + \beta), cos⁡α\cos \alpha, and cos⁡β\cos \beta.
  4. Conclude that the identity cos⁡(α+β)=cos⁡α⋅cos⁡β−sin⁡α⋅sin⁡β\cos(\alpha + \beta) = \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta holds true.