The minimum value of 3cosx + 4sinx + 8 is ______.

प्रश्न

विकल्प

MCQ

रिक्त स्थान भरें

उत्तर

The minimum value of 3cosx + 4sinx + 8 is 3.

Explanation:

The given expression is 3cosx + 4sinx + 8

Let y = 3cosx + 4sinx + 8

⇒ y – 8 = 3cosx + 4sinx

Minimum value of y – 8 = `sqrt((3)^2 + (4)^2`

⇒ y – 8 = `-sqrt(9 + 16)` = – 5

⇒ y = 8 – 5 = 3

So, the minimum value of the given expression is 3.

shaalaa.com

Trigonometric Equations

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 41 Q 40 Q 42

अध्याय 3: Trigonometric Functions - Exercise [पृष्ठ ५६]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11

अध्याय 3 Trigonometric Functions
Exercise | Q 41 | पृष्ठ ५६

वीडियो ट्यूटोरियलVIEW ALL [1]

view
वीडियो ट्यूटोरियल For All Subjects
Trigonometric Equations

video tutorial00:19:05

संबंधित प्रश्न

Find the general solution for each of the following equations sec² 2x = 1– tan 2x

If \[\tan x = \frac{a}{b},\] show that

\[\frac{a \sin x - b \cos x}{a \sin x + b \cos x} = \frac{a^2 - b^2}{a^2 + b^2}\]

Prove that:

\[\sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6} = \frac{1}{2}\]

Prove that:

\[3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1\]

In a ∆ABC, prove that:

\[\cos\left( \frac{A + B}{2} \right) = \sin\frac{C}{2}\]

In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0

Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]

Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]

Prove that:

\[\tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} = 0\]

If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

The value of sin²5° + sin²10° + sin²15° + ... + sin²85° + sin²90° is

If x sin 45° cos² 60° = \[\frac{\tan^2 60^\circ cosec30^\circ}{\sec45^\circ \cot^{2^\circ} 30^\circ}\], then x =

Find the general solution of the following equation:

\[cosec x = - \sqrt{2}\]

Find the general solution of the following equation:

\[\sin 2x = \frac{\sqrt{3}}{2}\]

Find the general solution of the following equation:

\[\tan 3x = \cot x\]