Trigonometry Proof: Expression Independent of Theta

This lesson proves an important class 10 trigonometry result: that the expression below has the same value for every angle $\theta$. We expand each term with algebraic identities, use $\sin^2\theta + \cos^2\theta = 1$, and watch the angle-dependent parts cancel.

The expression to evaluate

We want to show that

$3(\sin\theta - \cos\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4(\sin^6\theta + \cos^6\theta)$

is independent of $\theta$.

Rewriting each term

We first rewrite the powers so we can apply standard identities:

• $(\sin\theta - \cos\theta)^4 = \big[(\sin\theta - \cos\theta)^2\big]^2$
• $\sin^6\theta + \cos^6\theta = (\sin^2\theta)^3 + (\cos^2\theta)^3$, a sum of cubes.

First term

Using $(a - b)^2 = a^2 - 2ab + b^2$ with the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:

$(\sin\theta - \cos\theta)^2 = 1 - 2\sin\theta\cos\theta$

Squaring again,

$(\sin\theta - \cos\theta)^4 = (1 - 2\sin\theta\cos\theta)^2 = 1 - 4\sin\theta\cos\theta + 4\sin^2\theta\cos^2\theta$

Multiplying by $3$:

$3(\sin\theta - \cos\theta)^4 = 3 - 12\sin\theta\cos\theta + 12\sin^2\theta\cos^2\theta$

Second term

Using $(a + b)^2 = a^2 + 2ab + b^2$:

$(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta$

Multiplying by $6$:

$6(\sin\theta + \cos\theta)^2 = 6 + 12\sin\theta\cos\theta$

Third term

Using the sum of cubes $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ with $a = \sin^2\theta$, $b = \cos^2\theta$, so that $a + b = 1$:

$\sin^6\theta + \cos^6\theta = 1^3 - 3\sin^2\theta\cos^2\theta \cdot 1 = 1 - 3\sin^2\theta\cos^2\theta$

Multiplying by $4$:

$4(\sin^6\theta + \cos^6\theta) = 4 - 12\sin^2\theta\cos^2\theta$

Adding everything together

Lining up the three expanded terms:

$\big(3 - 12\sin\theta\cos\theta + 12\sin^2\theta\cos^2\theta\big) + \big(6 + 12\sin\theta\cos\theta\big) + \big(4 - 12\sin^2\theta\cos^2\theta\big)$

The $\sin\theta\cos\theta$ terms cancel ($-12$ and $+12$), and the $\sin^2\theta\cos^2\theta$ terms cancel ($+12$ and $-12$). Only the constants remain:

$3 + 6 + 4 = 13$

Since no $\theta$ survives, the value is always $13$.

Key takeaways

• $(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta$, straight from the Pythagorean identity.
• $\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$ via the sum of cubes.
• After expansion, every angle-dependent term cancels, leaving the constant $13$, so the expression is independent of $\theta$.