"For 0o ≤ x ≤ 180o, find to 3 significant figures the range of values of x for: 5 sin x > cos x."
In an earlier post, I have blogged about "A Negative Number Without The Negative Sign". In that post, I blogged about log x which is -ve but is not showing the -ve sign (log x, where 0 < x < 1). And, at the end of that post, I invited readers to think about other type(s) of -ve numbers without the -ve signs.
Have you thought about it? That depends on whether you are 'passionate' about that question, right? Any way, let me continue: Trigonometric ratios are, among others, such numbers: -ve numbers which may not show -ve signs.
Depending on the quadrants that the angles fall into, trigonometric ratios can be +ve or -ve. For examples:
If you are trying to solve the above inequality algebraically, you can approach it quadrant by qaudrant. Dividing by cosine in the 2nd qd is dividing by a -ve number and, the inequality sign must be reversed!
However, you can also solve the above inequality by sketching the graphs of y = 5 sin x and y = cos x to identify y of the 1st graph that is above (>) the y of the 2nd graph, then use 5 sin x = cos x to find the root to determine bound of the range of values of x -- much in the same way as you would do for quadratic inequalities that are formed from the nature of the quadratic roots and the discriminant (2 equal roots means b2 – 4ac = 0; 2 unequal roots means b2 – 4ac > 0; or, no real roots means b2 – 4ac < 0).
Okay, you may compare your answer with mine by posting a comment to this post.
Have a productive day! (My answer: 11.3o < x ≤ 180o)
In an earlier post, I have blogged about "A Negative Number Without The Negative Sign". In that post, I blogged about log x which is -ve but is not showing the -ve sign (log x, where 0 < x < 1). And, at the end of that post, I invited readers to think about other type(s) of -ve numbers without the -ve signs.
Have you thought about it? That depends on whether you are 'passionate' about that question, right? Any way, let me continue: Trigonometric ratios are, among others, such numbers: -ve numbers which may not show -ve signs.
Depending on the quadrants that the angles fall into, trigonometric ratios can be +ve or -ve. For examples:
- In the 1st quadrant (qd), ALL trigo ratios are +ve; ---------All ~ Add
- In the 2nd qd, only sines (and thus, cosec) are +ve; --------Sine ~ Sugar
- In the 3rd qd, only tangents (and thus, cot) are +ve; and ---Tangent ~ To
- In the 4th qd, only cosines (and, sec) are +ve --------------Cosine ~ Coffee
- My mnemonic for students to remember the above: Add Sugar To Coffee (since this post is written from TTDI Starbucks, :))
If you are trying to solve the above inequality algebraically, you can approach it quadrant by qaudrant. Dividing by cosine in the 2nd qd is dividing by a -ve number and, the inequality sign must be reversed!
However, you can also solve the above inequality by sketching the graphs of y = 5 sin x and y = cos x to identify y of the 1st graph that is above (>) the y of the 2nd graph, then use 5 sin x = cos x to find the root to determine bound of the range of values of x -- much in the same way as you would do for quadratic inequalities that are formed from the nature of the quadratic roots and the discriminant (2 equal roots means b2 – 4ac = 0; 2 unequal roots means b2 – 4ac > 0; or, no real roots means b2 – 4ac < 0).
Okay, you may compare your answer with mine by posting a comment to this post.
Have a productive day! (My answer: 11.3o < x ≤ 180o)
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