Download App

Transformation Formulae — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae5 Marks
Question
Prove that:
$\cos(\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})\\+\cos(-\text{A+B+C})=4\cos\text{A}\cos\text{B}\cos\text{C}$
✓

Answer

We have,
$\text{LHS}=\cos(\text{A+B+C})+\cos(\text{A}-\text{B}+\text{C})\\ \ \ \ \ +\cos(\text{A+B}-\text{C})+\cos(-\text{A+B+C})$
$=\ [\cos(\text{A+B+C})+\cos(\text{A}-\text{B+C})]\\ \ \ \ \ +[\cos(\text{A+B}-\text{C})+\cos(-\text{A+B+C})$
$=\ 2\cos\Big\{\frac{\text{A+B+C+A}-\text{B}+\text{C}}{2}\Big\}\cos\Big\{\frac{\text{A+B+C}-\text{A+B}-\text{C}}{2}\Big\}\\ \ \ \ +2\begin{Bmatrix}\cos\Big\{\frac{\text{A+B}-\text{C}-\text{A+B+C}}{2}\Big\} \\\cos\Big\{\frac{\text{A+B}-\text{C}+\text{A}-\text{B}-\text{C}}{2}\Big\} \end{Bmatrix}$
$=\ 2\cos\Big(\frac{2\text{A}+2\text{C}}{2}\Big)\cos\Big(\frac{2\text{B}}{2}\Big)+2\cos\Big(\frac{2\text{B}}{2}\Big)\cos\Big\{\frac{2\text{A}-2\text{C}}{2}\Big\}$
$=\ 2\cos(\text{A+B})\cos(\text{B})+2\cos(\text{B})\cos(\text{A}-\text{C})$
$=\ 2\cos(\text{B})[\cos(\text{A+C})+\cos(\text{A}-\text{C})]$
$=\ 2\cos\text{B}\Big[2\cos\Big(\frac{\text{A+C+A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}-\text{A+C}}{2}\Big)\Big]$
$=\ 2\cos(\text{B})[2\cos\text{A}\cos\text{C}]$
$=\ 4\cos\text{A}\cos\text{B}\cos{C}.$
$=\ \text{RHS}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Transformation Formulae→Transformation Formulae — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

If $\theta_1,\ \theta_2,\ \theta_3,\ ...\theta_\text{n}$ are in AP. whose common difference is d, show thet $\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+...+\sec\theta_{\text{n}-1}\sec\theta_\text{n}=\frac{\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$
→
$P_1, P_2$ are points on either of the two lines $\text{y} - \sqrt{3}|\text{x}| = 2$ at a distance of $5$ units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from$ P1 , P2$ on the bisector of the angle between the given lines.
$[$Hint: Lines are $\text{y} = \sqrt{3}\text{x} + 2$ and $\text{y} = -\sqrt{3}\text{x} + 2$ according as $\text{x} \geq 0$ or $x < 0. y-$ axis is the bisector of the angles between the lines. $P_1, P_2$ are the points on these lines at a distance of $5$ units from the point of intersection of these lines which have a point on $y-$axis as common foot of perpendiculars from these points. The y$-$coordinate of the foot of the perpendicular is given by $2 + 5 \cos30^\circ.]$
→
A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has:
  1. No girl?
  2. At least one boy and one girl?
  3. At least $3$ girls?
→
Find the number of words formed by permuting all the letters of the following words:
PAKISTAN.
→
Prove the following statement by principle of mathematical induction:
$2 + 4 + 6 + ..... + 2n = n^2 + n$ for all natural numbers $n.$
→
Find the distance from the eye at which a coin of 2cm diameter should be held so as to conceal the full moon whose angular diameter is 31'.
→
Prove that $\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}=\frac{1}{16}$
→
Show that the solution set of the following system of linear inequalities is an unbounded region: 2x + y > 8, x + 2y > 10, x > 0, y > 0.
→
If $0\leq\text{x}\leq\pi$ and x lies in the IInd quadrant such that $\sin\text{x}=\frac{1}{4}.$ Find the values of $\cos\frac{\text{x}}{2},\sin\frac{\text{x}}{2}$ and $\tan\frac{\text{x}}{2}$
→
$\frac{\sqrt{\sin\text{A}}-\sqrt{\sin\text{B}}}{\sqrt{\sin\text{A}}+\sqrt{\sin\text{B}}}=\frac{\text{a + b}-2\sqrt{\text{ab}}}{\text{a}-\text{b}}$
→