Because sinθ and cosθ are just the legs of a right triangle scaled to a hypotenuse of 1, they can never run away to infinity — they are boxed in. That single fact is what every “maximum & minimum” question leans on: find the box, and the answer is one of its edges.
The bounds that decide every answer
• Plain sine and cosine are trapped:−1≤sinθ≤1 and −1≤cosθ≤1. So sinθ peaks at 1 and bottoms at −1.
• Mix of sine and cosine: the value of asinθ+bcosθ always lies in [−a2+b2,a2+b2]. So its maximum is a2+b2 and its minimum is −a2+b2.
• Weighted squares: since sin2θ+cos2θ=1, the value of asin2θ+bcos2θ just slides between a and b — the answer is simply the smaller one (min) or the larger one (max).
• Reciprocal pairs (AM-GM): for shapes like acsc2θ+bsin2θ, acot2θ+btan2θ or asec2θ+bcsc2θ, the two terms multiply to a number free of θ, so the minimum is 2ab — provided the equalising angle is legal (sin2θ≤1). If not, the minimum is just a+b.
📐Right Triangle ExplorerPick an angle — see exact fraction values
Select angle θ
sin θP / H1/2
cos θB / H√3/2
tan θP / B1/√3
cosec θH / P2
sec θH / B2/√3
cot θB / P√3
Try it: the maximum value of 3sinθ+4cosθ is 32+42=25= .
📝Practice Questions
Q1The maximum value of 5sinθ + 12cosθ is:
Q2The minimum value of sec²θ + cosec²θ is:
Q3The minimum value of 4cosec²θ + 9sin²θ is:
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Real exam questions — Maximum & Minimum
1 question types · 5 solved examples from real SSC papers
Maximum and minimum questions reward one habit: turn the expression into a reciprocal pair and read off the answer with AM-GM instead of testing angles one by one.
How to solve this type
In each of these two terms are reciprocal partners, so their product loses θ entirely. For any two positives p+q≥2pq, with equality when p=q, so the minimum is 2ab — but only if the angle that makes the terms equal is legal (sin2θ≤1). For asin2θ+bcsc2θ with the cosec coefficient bigger (b>a) that angle is impossible, so the minimum jumps to a+b at θ=90∘. For asec2θ+bcsc2θ, split off the constants first to land on (a+b)2.
