If - a sin 2- b sin-cos- c cos 2-1-2 a - c - -1-2 k- then k 2 is equal toA- b2-a-c2B- a2-b-c2C- c2-a-b2D- None of these

The correct option is A a sin^2 θ+b sin θ cos θ+c cos^2 θ 1/2(a+c) =1/2[ acos 2θ + b sin 2θ + c cos 2 θ] =1/2[b sin 2 θ (a c)cos 2 θ] ∵ |b sin 2 θ (a c)co ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If | a sin 2θ...

Question

$I f ∣ ∣ a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) ∣ ∣ \leq \frac{1}{2} k, t h e n k^{2} i s e q u a l t o$

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Solution

The correct option is A
$a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) = \frac{1}{2} [- a c o s 2 θ + b s i n 2 θ + c c o s 2 θ] = \frac{1}{2} [b s i n 2 θ - (a - c) c o s 2 θ] ∵ | b s i n 2 θ - (a - c) c o s 2 θ | \leq \sqrt{b^{2} + (a - c)^{2}} ∴ ∣ ∣ \frac{1}{2} {b s i n 2 θ - (a - c) c o s 2 θ} ∣ ∣ \leq \frac{1}{2} \sqrt{b^{2} + (a - c)^{2}} \Rightarrow ∣ ∣ a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) ∣ ∣ \leq \frac{1}{2} \sqrt{b^{2} + (a - c)^{2}}$

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Similar questions

Q. If

∣ ∣ ∣ a {sin}^{2} θ + b sin θ cos θ + c {cos}^{2} θ - \frac{1}{2} (a + c) ∣ ∣ ∣ \leq \frac{1}{2} k

, then

k^{2}

is equal to

Q. If a cos θ − b sin θ = c, then a sin θ + b cos θ =

(a)

\pm \sqrt{a^{2} + b^{2} + c^{2}}

(b)

\pm \sqrt{a^{2} + b^{2} - c^{2}}

(c)

\pm \sqrt{c^{2} - a^{2} - b^{2}}

(d) None of these

Find:
$(a + b + c) \times (a + b - c) =$

Q. The distance of the point P (a, b, c) from the x-axis is
(a)

\sqrt{b^{2} + c^{2}}

(b)

\sqrt{a^{2} + c^{2}}

(c)

\sqrt{a^{2} + b^{2}}

(d) none of these

a² + b² + c² - ab - bc - ca equals:

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If∣∣a sin2 θ+b sin θ cos θ+c cos2 θ−12(a+c)∣∣≤12k, then k2 is equal to

$I f ∣ ∣ a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) ∣ ∣ \leq \frac{1}{2} k, t h e n k^{2} i s e q u a l t o$