(a) Draw the graph of y = 1/√2 (sin x + cos x) from x = -π/2 to x = π/2. (b) If cos(α + β) = 4/5, si

Q: Wait, why do we reject the negative root when finding $\cos(\alpha - \beta)$? How do we know it is in the first quadrant?

Great observation! We are given that $0 < \alpha, \beta < \frac{\pi}{4}$. This implies that the difference $(\alpha - \beta)$ must lie between $-\frac{\pi}{4}$ and $\frac{\pi}{4}$. However, we are given that $\sin(\alpha - \beta) = \frac{5}{13}$, which is a positive value. Since the sine is positive, the angle must be in the first quadrant (between $0$ and $\frac{\pi}{4}$), not the fourth. Therefore, the cosine must also be positive.

Q: Can't we just solve for $\alpha$ and $\beta$ using inverse trigonometric functions?

Technically, yes, but you are walking into a minefield. Calculating $\alpha = \arcsin(...)$ will lead to non-standard angles that are extremely difficult to work with algebraically. In JEE Advanced, we value elegance and efficiency. Using the identity $2\alpha = (\alpha + \beta) + (\alpha - \beta)$ is the intended path because it keeps the math clean and minimizes the chance of calculation errors.

Q: Why did we convert the function to $\sin(x + \frac{\pi}{4})$ instead of just plotting points for the original equation?

You could plot points, but that is the 'brute force' method. By simplifying to $\sin(x + \frac{\pi}{4})$, you immediately understand the physics of the graph: it is a standard sine wave shifted to the left by $\frac{\pi}{4}$. This conceptual understanding allows you to identify the maximum, minimum, and intercepts instantly without needing to calculate a table of values.

Question

(a) Draw the graph of $y = \frac{1}{\sqrt{2}} (\sin x + \cos x)$ from $x = -\pi/2$ to $x = \pi/2$. (b) If $\cos(\alpha + \beta) = 4/5, \sin(\alpha - \beta) = 5/13$, and $\alpha, \beta$ lies between 0 and $\pi/4$, find $	an 2\alpha$.

Sigma Percentile · Accepted Answer

### Analyzing the Trigonometric Wave

To simplify the expression $y = \frac{1}{\sqrt{2}}(\sin x + \cos x)$, we look for a hidden harmony within the sum. By distributing the constant, we rewrite the function as:

$$y = \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}}$$

Recognizing that $\frac{1}{\sqrt{2}} = \cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4})$, we apply the sine addition formula, $\sin(A+B) = \sin A \cos B + \cos A \sin B$. This transforms our expression into:

$$y = \sin\left(x + \frac{\pi}{4}ight)$$

This represents a standard sine wave shifted to the left by $\frac{\pi}{4}$. Over the interval $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the function maps the journey of the wave from $\sin(-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$ to $\sin(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}}$, reaching its peak at $x = \frac{\pi}{4}$.

### The Algebraic Dance: Solving for $	an 2\alpha$

We are given $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$, with $\alpha, \beta \in (0, \frac{\pi}{4})$. Rather than solving for individual angles, we express $2\alpha$ as the sum of the given arguments:

$$2\alpha = (\alpha + \beta) + (\alpha - \beta)$$

Let $A = \alpha + \beta$ and $B = \alpha - \beta$. Given the constraints, both $A$ and $B$ lie in the first quadrant. We calculate their tangents as follows:

For $A$: Since $\cos A = \frac{4}{5}$, then $\sin A = \frac{3}{5}$, yielding $	an A = \frac{3}{4}$.

For $B$: Since $\sin B = \frac{5}{13}$, then $\cos B = \frac{12}{13}$, yielding $	an B = \frac{5}{12}$.

### Final Calculation

We now apply the tangent addition formula, $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$, to find $	an 2\alpha$:

$$	an 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \left(\frac{3}{4} \cdot \frac{5}{12}ight)}$$

Simplifying the numerator and denominator:

$$	ext{Numerator: } \frac{9+5}{12} = \frac{14}{12} = \frac{7}{6}$$

$$	ext{Denominator: } 1 - \frac{15}{48} = 1 - \frac{5}{16} = \frac{11}{16}$$

Combining these results, we obtain the final value:

$$	an 2\alpha = \frac{7}{6} \cdot \frac{16}{11} = \mathbf{\frac{56}{33}}$$

Animated Solution for Mathematics - Trigonometry: (a) Draw the graph of $y = \frac{1}{2} (sin x + cos x)$ from $x = - π /2$ to $x = π /2$ . (b) If $cos (α + β) = 4/5, sin (α - β) = 5/13$ , and $α, β$ lies between 0 and $π /4$ , find $tan 2 α$ .

Visualized Solution

Simplifying the Function $y$

Evaluating at the Boundaries

Evaluating at Critical Points

Drawing the Graph

Finding $tan (α + β)$

Finding $tan (α - β)$

The $2 α$ Identity Trick

Final Calculation

The Sigma Insight: Trigonometric Functions of Compound Angles

Analyzing the Trigonometric Wave

The Algebraic Dance

Final Calculation

Similar Questions

Let $cos (α + β) = 4/5$ and $sin (α - β) = 5/13$ , where $0 \leq α, β \leq π /4$ . Then $tan 2 α =$

If $cos (α + β) = \frac{3}{5}, sin (α - β) = \frac{5}{13}$ and $0 < α, β < \frac{π}{4}$ , then $tan (2 α)$ is equal to :

If $tan α = \frac{m}{m + 1}$ and $tan β = \frac{1}{2 m + 1}$ , find the possible values of $(α + β)$ .

If $cot α = 1$ and $sec β = - \frac{5}{3}$ , where $π < α < \frac{3 π}{2}$ and $\frac{π}{2} < β < π$ , then the value of $tan (α + β)$ and the quadrant in which $α + β$ lies, respectively are

If $tan A = \frac{1}{x ( x ^{2} + x + 1 )}$ , $tan B = \frac{x}{x ^{2} + x + 1}$ and $tan C = (x^{- 3} + x^{- 2} + x^{- 1})^{\frac{1}{2}}$ , $0 < A, B, C < \frac{π}{2}$ , then $A + B$ is equal to :

Determine the smallest positive value of $x$ (in degrees) for which $tan (x + 10 0^{\circ}) = tan (x + 5 0^{\circ}) tan x tan (x - 5 0^{\circ})$ .

In a triangle $P QR, ∠ R = π /2$ . If $tan (P /2)$ and $tan (Q /2)$ are the roots of $a x^{2} + b x + c = 0, a \neq = 0$ then

If the roots of the quadratic equation $x^{2} + p x + q = 0$ are $tan 3 0^{\circ}$ and $tan 1 5^{\circ}$ , respectively, then the value of $2 + q - p$ is

The value of the expression $3 csc 2 0^{\circ} - sec 2 0^{\circ}$ is equal to

If $f (x) = cos (ln x)$ , then $f (x) f (y) - \frac{1}{2} [f (x / y) + f (x y)]$ has the value

Visualized Solution

Simplifying the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Evaluating at the Boundaries

Evaluating at Critical Points

Drawing the Graph

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(α+β)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(α−β)

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2α Identity Trick

Final Calculation

The Sigma Insight: Trigonometric Functions of Compound Angles

Analyzing the Trigonometric Wave

The Algebraic Dance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let cos(α+β)=4/5 and sin(α−β)=5/13, where 0≤α,β≤π/4. Then tan2α=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If cos(α+β)=53​,sin(α−β)=135​ and 0<α,β<4π​, then tan(2α) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If tanα=m+1m​ and tanβ=2m+11​, find the possible values of (α+β).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If cotα=1 and secβ=−35​, where π<α<23π​ and 2π​<β<π, then the value of tan(α+β) and the quadrant in which α+β lies, respectively are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If tanA=x(x2+x+1)​1​, tanB=x2+x+1​x​​ and tanC=(x−3+x−2+x−1)21​, 0<A,B,C<2π​, then A+B is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determine the smallest positive value of x (in degrees) for which tan(x+100∘)=tan(x+50∘)tanxtan(x−50∘).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle PQR,∠R=π/2. If tan(P/2) and tan(Q/2) are the roots of ax2+bx+c=0,a=0 then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the roots of the quadratic equation x2+px+q=0 are tan30∘ and tan15∘, respectively, then the value of 2+q−p is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the expression 3​csc20∘−sec20∘ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=cos(lnx), then f(x)f(y)−21​[f(x/y)+f(xy)] has the value

Simplifying the Function $y$

Finding $tan (α + β)$

Finding $tan (α - β)$

The $2 α$ Identity Trick

Let $cos (α + β) = 4/5$ and $sin (α - β) = 5/13$ , where $0 \leq α, β \leq π /4$ . Then $tan 2 α =$

If $cos (α + β) = \frac{3}{5}, sin (α - β) = \frac{5}{13}$ and $0 < α, β < \frac{π}{4}$ , then $tan (2 α)$ is equal to :

If $tan α = \frac{m}{m + 1}$ and $tan β = \frac{1}{2 m + 1}$ , find the possible values of $(α + β)$ .

If $cot α = 1$ and $sec β = - \frac{5}{3}$ , where $π < α < \frac{3 π}{2}$ and $\frac{π}{2} < β < π$ , then the value of $tan (α + β)$ and the quadrant in which $α + β$ lies, respectively are

If $tan A = \frac{1}{x ( x ^{2} + x + 1 )}$ , $tan B = \frac{x}{x ^{2} + x + 1}$ and $tan C = (x^{- 3} + x^{- 2} + x^{- 1})^{\frac{1}{2}}$ , $0 < A, B, C < \frac{π}{2}$ , then $A + B$ is equal to :

Determine the smallest positive value of $x$ (in degrees) for which $tan (x + 10 0^{\circ}) = tan (x + 5 0^{\circ}) tan x tan (x - 5 0^{\circ})$ .

In a triangle $P QR, ∠ R = π /2$ . If $tan (P /2)$ and $tan (Q /2)$ are the roots of $a x^{2} + b x + c = 0, a \neq = 0$ then

If the roots of the quadratic equation $x^{2} + p x + q = 0$ are $tan 3 0^{\circ}$ and $tan 1 5^{\circ}$ , respectively, then the value of $2 + q - p$ is

The value of the expression $3 csc 2 0^{\circ} - sec 2 0^{\circ}$ is equal to

If $f (x) = cos (ln x)$ , then $f (x) f (y) - \frac{1}{2} [f (x / y) + f (x y)]$ has the value