Question

If cos theta - 4 sin theta = 1, then sin + 4 cos theta =​

Answer 1

Given:

• Cos theta - 4 sin theta = 1

To find:

• Sin + 4 cos theta = ?

Solution:

Cos theta - 4 sin theta = 1 — Eq - 1...

Sin theta + 4 cos theta = x — Eq - 2...

Squaring and adding both equations :

(Cos theta)² - (4 sin theta)² + (Sin theta)² + (4 cos theta)² = 1² + x²

→ Cos² theta + 16 Sin² theta + Sin² theta + 16 cos² theta = 1 + x²

→ Cos² theta + Sin² theta + 16 cos² theta + 16 Sin² theta = 1 + x²

★ Sin² theta + Cos² theta = 1

→ 1 + 16 cos² theta + 16 Sin² theta = 1 + x²

Taking 16 common,

→ 1 + 16 (cos² theta + Sin² theta) = 1 + x²

→ 1 + 16 (1) = 1 + x²

→ 1 + 16 = 1 + x²

→ 1 - 1 + 16 = x²

→ 16 = x²

→ x = √16

→ x = ±4

∴ Hence, sin + 4 cos theta = ±4$.$

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More info :

Trigonometric Identities :

• sin²θ + cos²θ = 1
• sec²θ - tan²θ = 1
• csc²θ - cot²θ = 1

Trigonometric relations :

• sinθ = 1/cscθ
• cosθ = 1 /secθ
• tanθ = 1/cotθ
• tanθ = sinθ/cosθ
• cotθ = cosθ/sinθ

Answer 2

Given:-

$\red{➤}\:$$\sf \cos \theta - 4 \sin \theta = 1$

To Find:-

$\orange{☛}\:$$\sf Value\:of\: \sin \theta + 4 \cos \theta$

Solution:-

$\begin{gathered}\\\quad\longrightarrow\quad\sf \cos \theta - 4 \sin \theta = 1\quad\quad(given) \\\end{gathered}$

Squaring Both Sides-

$\begin{gathered}\\\quad\longrightarrow\quad\sf (\cos \theta - 4 \sin \theta)² = 1² \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf {\cos}^2 \theta + 16 {\sin}^2 \theta-2. \sin\theta.4\cos\theta = 1\quad(a-b)²=a²+b²-2ab)\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf 8\sin\theta\cos\theta ={\cos}^2 \theta + 16 {\sin}^2 \theta-1 \_ \:\_ \:\_ \:(1)\\\end{gathered}$

Now,

$\begin{gathered}\\\quad\longrightarrow\quad \sf (sin\theta + 4\cos\theta)² = sin²\theta + 16\cos²\theta + 8\cos\theta\sin\theta\\\end{gathered}$

Substituting (1)-

$\begin{gathered}\\\quad\longrightarrow\quad \sf (sin\theta + 4\cos\theta)² = sin²\theta + 16\cos²\theta +{\cos}^2 \theta + 16 {\sin}^2 \theta-1 \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf (sin\theta + 4\cos\theta)² = sin²\theta +{\cos}^2 \theta-1 + 16\cos²\theta + 16 {\sin}^2 \theta \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf (sin\theta + 4\cos\theta)² = 1-1 + 16(\cos²\theta + {\sin}^2 \theta)\quad (sin^2\theta + cos^2\theta = 1) \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf (sin\theta + 4\cos\theta)² = 16\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf (sin\theta + 4\cos\theta)² = 4²\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \boxed{\sf (sin\theta + 4\cos\theta) = ± 4}\\\end{gathered}$