For all θ in [0, π/2] show that, cos(sin θ) ≥ sin(cos θ).

Q: Wait, why do we flip the inequality sign when removing the cosine?

Great catch! This is a classic trap. The function $f(x) = \cos x$ is strictly decreasing on the interval $[0, \frac{\pi}{2}]$. For any strictly decreasing function, if $f(A) \ge f(B)$, it implies that $A \le B$. The inequality sign must flip because as the input increases, the output decreases.

Q: Why do we need to check if the arguments are in $[0, \frac{\pi}{2}]$?

We must ensure the arguments lie within the interval where the cosine function is strictly decreasing. If an argument fell outside this range, the monotonicity property might not hold, and our logic would fail. We verified that $\sin \theta \in [0, 1]$ and $\frac{\pi}{2} - \cos \theta \in [\frac{\pi}{2} - 1, \frac{\pi}{2}]$, both of which are within $[0, \frac{\pi}{2}]$.

Q: Can we use calculus instead of the $a \sin \theta + b \cos \theta$ formula?

Absolutely! You could define $f(\theta) = \sin \theta + \cos \theta$, find its derivative $f'(\theta) = \cos \theta - \sin \theta$, and set it to zero to find the critical point at $\theta = \frac{\pi}{4}$. Evaluating the function at this point gives the maximum value of $\sqrt{2}$. Both methods are perfectly valid and lead to the same conclusion.

Question

For all $	heta$ in $[0, \pi/2]$ show that, $\cos(\sin 	heta) \ge \sin(\cos 	heta)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the edge of a vast mathematical landscape, looking at the expression $\cos(\sin 	heta) \ge \sin(\cos 	heta)$. It looks intimidating, but it is a beautiful puzzle waiting to be solved.

Our goal is to prove this inequality for all $	heta$ in the first quadrant, $	heta \in [0, \frac{\pi}{2}]$.

### The Transformation

The first step is to make the outer functions identical. We cannot easily compare a cosine to a sine, so we use the complementary angle identity: $\sin x = \cos(\frac{\pi}{2} - x)$.

By substituting $x = \cos 	heta$, we transform the right-hand side:
$$\sin(\cos 	heta) = \cos\left(\frac{\pi}{2} - \cos 	hetaight)$$

Now, our inequality looks much more manageable:
$$\cos(\sin 	heta) \ge \cos\left(\frac{\pi}{2} - \cos 	hetaight)$$

### The Trap of Monotonicity

Many students stumble here by simply 'canceling' the cosines. However, the function $f(x) = \cos x$ is strictly decreasing on $[0, \frac{\pi}{2}]$.

This means that if $\cos A \ge \cos B$, then $A \le B$. The inequality sign flips.

Before proceeding, we must verify that both arguments, $A = \sin 	heta$ and $B = \frac{\pi}{2} - \cos 	heta$, lie within the interval $[0, \frac{\pi}{2}]$. For $	heta \in [0, \frac{\pi}{2}]$, $\sin 	heta$ ranges from $0$ to $1$, which is less than $\frac{\pi}{2} \approx 1.57$.

Similarly, $\cos 	heta$ ranges from $0$ to $1$, so $\frac{\pi}{2} - \cos 	heta$ ranges from $\frac{\pi}{2} - 1$ to $\frac{\pi}{2}$. Both arguments are safe, so we can drop the cosines and flip the sign:
$$\sin 	heta \le \frac{\pi}{2} - \cos 	heta$$

### The Final Showdown

We have arrived at a much simpler inequality:
$$\sin 	heta + \cos 	heta \le \frac{\pi}{2}$$

To prove this holds for all $	heta$, we find the maximum value of the function $f(	heta) = \sin 	heta + \cos 	heta$. Recall the standard algebraic result: the maximum value of $a \sin 	heta + b \cos 	heta$ is $\sqrt{a^2 + b^2}$.

Here, $a=1$ and $b=1$, so the maximum value is:
$$\sqrt{1^2 + 1^2} = \sqrt{2}$$

This peak occurs at $	heta = \frac{\pi}{4}$. Now, we compare this maximum value, $\sqrt{2} \approx 1.414$, with the right-hand side, $\frac{\pi}{2} \approx 1.571$.

Since $1.414 < 1.571$, the function $\sin 	heta + \cos 	heta$ never exceeds $\frac{\pi}{2}$. We have successfully proven that **$\cos(\sin 	heta) \ge \sin(\cos 	heta)$** for all $	heta \in [0, \frac{\pi}{2}]$.

Animated Solution for Mathematics - Trigonometry: For all $θ$ in $[0, π /2]$ show that, $cos (sin θ) \geq sin (cos θ)$ .

Visualized Solution

Objective and Setup

Complementary Angle Identity

Transforming the Inequality

Analyzing the Arguments

Bounding the First Argument

Bounding the Second Argument

Applying the Decreasing Property

Rearranging the Terms

Maximum Value of $a sin θ + b cos θ$

Calculating the Maximum Value

Comparing with $\frac{π}{2}$

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Transformation

The Trap of Monotonicity

The Final Showdown

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For all θ in [0,π/2] show that, cos(sinθ)≥sin(cosθ).