Trigonometric Functions - Class 11 Mathematics - Chapter 3 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Trigonometric Functions | NCERT | Mathematics | Class 11
If $\frac{\sin^{4} A}{a} + \frac{\cos^{4} A}{b} = \frac{1}{a + b'}$ then the value of $\frac{\sin^{8} A}{a^{3}} + \frac{\cos^{8} A}{b^{3}}$ is equal to
(A) $\frac{1}{(a + b)^{3}}$
B $\frac{a^{3}b^{3}}{(a + b)^{3}}$
(c) $\frac{a^2}{b^2}{(a + b)}^{2}$
D None of these
The correct option is A$$ \frac{1}{(a+b)^3} $$
It is initially provided that: $$ \frac{\sin^4 A}{a} + \frac{\cos^4 A}{b} = \frac{1}{a+b} $$ Expanding and simplifying the given equation leads to: $$ \frac{{(1 - \cos 2A)^2}}{4a} + \frac{(1 + \cos 2A)^2}{4b} = \frac{1}{a+b} $$ $$ \Rightarrow b(a+b)(1-2\cos 2A + \cos^2 2A) + a(a+b)(1 + 2\cos 2A + \cos^2 2A) = 4ab $$ Further simplify the equation: $$ (a+b)^2 \cos^2 2A + 2(a+b)(a-b) \cos 2A + (a-b)^2 = 0 \ \text{leads to } {(a+b) \cos 2A + (a-b)}^2 = 0 \text{ or } \cos 2A = \frac{b-a}{b+a} $$ Substituting $b=0$ based on the simplified relations, calculate the required expression: $$ \frac{\sin^8 A}{a^3} + \frac{\cos^8 A}{b^3} = \frac{(1-\cos 2A)^4}{16 a^3} + \frac{(1+\cos 2A)^4}{16 b^3} $$ Further simplifying: $$ = \frac{1}{16 a^3} \left[1 - \frac{b-a^3}{b+a}\right]^4 + \frac{1}{16 b^3} \left[1 + \frac{b-a^4}{b+a}\right]^4 \ = \frac{16 a^4}{16 a^3 (b+a)^4} + \frac{16 b^4}{16 b^3 (b+a)^4} \ = \frac{1}{(b+a)^4} (a+b) = \frac{1}{(a+b)^3} $$
Trick: Consider $\boldsymbol{A = 90^\circ}$, then: $$ \frac{\sin^8 A}{a^3} + \frac{\cos^8 A}{b^3} = \frac{1}{a^3} $$ which agrees with option (A). This result can be checked using other values of $\boldsymbol{A}$ as well.
$\lim_{x \to 0} \frac{3\sin^2 x - 2\sin x^2}{3x^2}$
To evaluate the given limit: $$ \lim_{x \to 0} \frac{3\sin^2 x - 2\sin x^2}{3x^2} $$ we can use the standard trigonometric limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and apply it appropriately.
The expression can be restructured into two separate limits: $$ \lim_{x \to 0} \frac{3\sin^2 x - 2\sin x^2}{3x^2} = \lim_{x \to 0} \frac{3\sin^2 x}{3x^2} - \lim_{x \to 0} \frac{2\sin x^2}{3x^2} $$
Each part can be analyzed as follows:
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The first part simplifies using the squaring property of the limit and $\frac{\sin x}{x} \to 1$ as $x \to 0$: $$ \lim_{x \to 0} \frac{3\sin^2 x}{3x^2} = \lim_{x \to 0}\left(\frac{\sin x}{x}\right)^2 = 1 $$
-
The second part, $\frac{\sin(x^2)}{x^2}$, needs a change of variable. Set $u = x^2$, as $x$ approaches $0$, $u$ also approaches $0$. Using this substitution: $$ \lim_{x \to 0} \frac{\sin x^2}{x^2} = \lim_{u \to 0} \frac{\sin u}{u} = 1 $$
Thus, substituting these limits back: $$ \lim_{x \to 0} \frac{3\sin^2 x}{3x^2} - \lim_{x \to 0} \frac{2\sin x^2}{3x^2} = 1 - \frac{2}{3}\times 1 = 1 - \frac{2}{3} = \frac{1}{3} $$
Therefore, the final answer is: $$ \boxed{\frac{1}{3}} $$
$\sin 90^{\circ} \cos 0^{\circ} + \sin 88^{\circ} \cos 2^{\circ} + \sin 86^{\circ} \cos 4^{\circ} + \ldots \ldots \ldots \ldots + \sin 2^{\circ} \cos 88^{\circ} + \sin 0^{\circ} \cos 90^{\circ}$
A) 1
B) 44
C) 22
D) 23
Correct Answer: D) 23
Explanation:
The given expression can be represented using the sum of angles formula, specifically identifying patterns convertible by angle sum identities: $$ \sin 90^\circ \cos 0^\circ + \sin 88^\circ \cos 2^\circ + \ldots + \sin 2^\circ \cos 88^\circ + \sin 0^\circ \cos 90^\circ $$
This series can be transformed utilizing the sum-to-product identities: $$ \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] $$ where $A + B = 90^\circ$ for the specific terms involved. Each term can be reduced using angle reduction identities: $$ \sin 90^\circ \cos 0^\circ + \sin 88^\circ \cos 2^\circ + \ldots + \sin 2^\circ \cos 88^\circ + \sin 0^\circ \cos 90^\circ = \cos(90^\circ - 90^\circ) + \cos(90^\circ - 88^\circ) + \ldots + \cos(90^\circ - 0^\circ) $$ This simplifies to: $$ \cos 0^\circ + \cos 2^\circ + \ldots + \cos 88^\circ $$
In angular terms, this spans angles at $2^\circ$ intervals, totaling 45 terms (including $0^\circ$ and $88^\circ$), however considering symmetry in cosine square values around $45^\circ$, expressions simplify to: $$ \cos^2 0^\circ + \cos^2 2^\circ + \ldots + \cos^2 44^\circ + \sin^2 44^\circ + \ldots + \sin^2 2^\circ + \sin^2 0^\circ $$
From 0° to 44° there are 23 terms including 0° itself, and the application of the fundamental identity of trigonometry: $$ \cos^2 \theta + \sin^2 \theta = 1 $$ tells us each pair sums to 1. Therefore, there are 23 terms: $$ 1 + 1 + \ldots + 1 \quad (\text{23 times}) $$
Sum is: $$ 23 $$
Hence, the correct answer is D) 23.
Let $f(n) = (\sin 1)(\sin 2)(\sin 3) \cdots (\sin n)$ for all $n \in \mathbb{N}$ where $n$ is in radians. Then the number of elements in the set $A = {f(1), f(2), \ldots, f(6)}$ that are positive is:
A) 2 B) 3 C) 4 D) 5
The question given asks for the number of elements in the set $$ A = {f(1), f(2), \ldots, f(6)} $$ where $$ f(n) = (\sin 1)(\sin 2)(\sin 3) \cdots (\sin n) $$ and each $n$ is in radians, that are positive.
We know:
- For $n = 1, 2, 3$, since $1$, $2$, and $3$ radians fall within the interval $(0, \pi)$, their sine values are positive.
- For $n = 4, 5, 6$, since $4$, $5$, and $6$ radians fall within the interval $(\pi, 2\pi)$, their sine values are negative.
Now, analyzing for each $f(n)$:
- $f(1) = \sin 1$ (positive)
- $f(2) = (\sin 1)(\sin 2)$ (positive as the product of two positive numbers is positive)
- $f(3) = (\sin 1)(\sin 2)(\sin 3)$ (positive as the product of three positive numbers is positive)
- $f(4) = (\sin 1)(\sin 2)(\sin 3)(\sin 4)$. Since $\sin 4$ is negative, the product turns negative.
- $f(5) = (\sin 1)(\sin 2)(\sin 3)(\sin 4)(\sin 5)$. Continuing with the product, the factor $\sin 5$ being negative will make this product positive again (as the product of an even number of negative numbers is positive).
- $f(6) = (\sin 1)(\sin 2)(\sin 3)(\sin 4)(\sin 5)(\sin 6)$. With $\sin 6$ also negative, the product now has three negative numbers, turning negative again.
Thus, $f(1), f(2), f(3), f(5)$ are positive. Counting these, we get 4 positive values.
The answer, therefore, is option $\mathbf{C} 4$.
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