Brain Teaser 3: Solve sin(3 – z) = 0.8 for 0 < z < π radians.

I have gone through this question with my students. You should try it too. You may post your answer by way of comments to this post with brief intro of yourself.

Either I or one of my students will put up the answer in a week or so. Have fun with brain teasers of this blog! :)

(Acknowledgment: The above question was taken from IGCSE Cambridge Maths 0606, Yr 2012 .... Q 9(b))

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My Proposed Solution:

For 0 < z < π rad, solve sin(3 – z) = 0.8

Get domain for (3 – z):

Since 0 < z < π rad, … x (-1) to get -z, reverse inequality sign

Therefore, 0 > -z > -π rad

+ 3:           3 > (3 – z) > 3 – π

                 3 > (3 – z) > 0.142…domain for principal angle obtained

Since, sin(3 – z) = 0.8,  and sin^-1(0.8) = 0.9272…(> 0.142)

therefore, (3 – z) is in 1^st quadrant (qd) or 2^nd qd:

1^st qd:      3 – z = sin^-1(0.8)

                     z = 3 - sin^-1(0.8) = 2.072704782

                     z = 2.07 (to 3 sf) (Answer)

2^nd qd:     3 – z = π - sin^-1(0.8) (< 3)

                      z = 3 - π + sin^-1(0.8) = 0.7857…

                      z = 0.786 (to 3 sf) (Answer)

Brain Teaser 2: Find the value of e^x, where x = 4 + ln 2

If you can solve the above, you should be able to solve these too (without using calculator):

Find the value of:
(a) 3^x, where x = 4 + log₃ 2. (My question)
(b) 2^z, where z = 5 + log₂ 3  (IGCSE Cambridge Add-Maths 0606, Yr 2010/Summer/p22, Q10(b))

I will post their answers in a week or so. Meanwhile, you may post your answers by way of comments to this post. Have fun with maths!

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29.04.2014

My Proposed Solutions to These Questions:

Find the value of:

 (Lead Question): e^x, where x = 4 + ln 2
(a) 3^x, where x = 4 + log₃ 2. (My question)
(b) 2^z, where z = 5 + log₂ 3  (IGCSE Cambridge Add-Maths 0606, Yr 2010/Summer/p22, Q10(b))

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For the lead question:

Since            x = 4 + ln 2,

therefore:     e^x = e^{(4 + ln 2)} = e⁴ x e^{ln 2} ..............Law of index: b^{(m + n)} = b^m x bⁿ

Now:           let y = e^{ln 2}

ln both sides: ln y = ln 2

therefore:          y = 2 = e^{ln 2} (Hence, remember: e^{ln a} = a; 10^{lg b} = b; etc.)

Hence, e^x = e^{(4 + ln 2)} = e⁴ x e^{ln 2} = 2e⁴ (Answer)

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(a)    Since           x = 4 + log₃ 2

therefore: 3^x = 3^(4 + log₃ 2) = 3⁴ x 3^log₃ 2 = 3⁴ x 2 (see above)

                  3^x = 81 x 2 = 162 (Answer)

(b)   Since         z = 5 + log₂ 3

therefore:  2^z = 2^(5 + log₂ 3) = 2⁵ x 2^log₂3 = 2⁵ x 3

                2^z   = 32 x 3 = 96 (Answer)

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So, remember:

·        e^{ln a} = a

·        10^{lg b} = b

·        b^log_b c = c

Have fun with maths!

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(1.5.2014)

Brain Teaser 2B (requires slightly more skill than Brain Teaser 2)

Find the exact value of:

1) 3^x , where x = 2 + log₉ 3

2) 2^y , where y = 3 + log₈ 2

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Brain Teaser 1: Trigonometric Inequality

"For 0^o ≤ x ≤ 180^o, 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:

• 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: 11.3^o < x ≤ 180^o)