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Confidence (out of 10) 15. A student attempts to answer the following question:...

Sep 24, 2024

Solution by Steps

step 1

Assume that \sin x - \cos x < 1 when

x

is an obtuse angle

step 2

This implies (\sin x - \cos x)^2 < 1

step 3

Expanding the left side gives \sin^2 x - 2\sin x \cos x + \cos^2 x < 1

step 4

Since

\sin^2 x + \cos^2 x = 1

, we can substitute to get 1 - 2\sin x \cos x < 1

step 5

Simplifying this leads to -2\sin x \cos x < 0

step 6

This implies \sin x \cos x > 0

step 7

For obtuse angles, \sin x > 0 and \cos x < 0 , thus \sin x \cos x < 0 , which contradicts our previous conclusion

step 8

Therefore, our assumption that \sin x - \cos x < 1 must be false, leading to

\sin x - \cos x \geq 1

Answer

The proof shows that

\sin x - \cos x \geq 1

for obtuse angles

x

Key Concept

Proof by contradiction is a method where we assume the opposite of what we want to prove and show that this leads to a contradiction.

Explanation

The contradiction arises from the properties of sine and cosine for obtuse angles, confirming that

\sin x - \cos x

must indeed be greater than or equal to 1.

Generate me a similar question

{'context': '[{\'Student\': \'my name is 123456789, and I am currently studying in College\', \'Sia\': \'awesome!\'}, {\'Student\': "\\nConfidence (out of 10)\\n12. A student attempts to answer the following question:\\n\\nGiven that

y

is an acute angle, use algebra to prove by contradiction that\\n\\\\[\\n\\\\tan y - 1 < 0\\n\\\\]\\n\\nThe student starts the proof with:\\n\\nAssume that $\\\\tan y - 1 \\geqslant 0$ when $y$ is an acute angle\\n\\\\[\\n\\\\begin{array}{l}\\n\\\\Rightarrow\\tan y \\geqslant 1 \\\\\\\\\\n\\\\Rightarrow \\\\ldots\\n\\\\end{array}\\n\\\\]\\n\\nThe start of the student\'s proof is reprinted below.\\nComplete the proof.\\n(3)\\n\\nAssume that $\\\\tan y - 1 \\geqslant 0$ when $y$ is an acute angle\\n\\\\[\\n\\\\Rightarrow\\tan y \\geqslant 1\\n\\\\]", \'Sia\': \' step 1 ⋮ Assume that \\\$ \\\\tan y - 1 \\geqslant 0 \\\$ when \\\$ y \\\$ is an acute angle. step 2 ⋮ This implies \\\$ \tan y \\geqslant 1 \\\\.\\ step 3 ⋮ Since \\\\( \\tan y = \\frac{\\sin y}{\\cos y} \\\$, we can rewrite the inequality. step 4 ⋮ This leads to \\\\( \\frac{\\sin y}{\\cos y} \\geqslant 1 \\\\. step 5 ⋮ This simplifies to \\\\( \\sin y \\geqslant \\cos y \\\$.\\n }]', 'question': ''}$