Monthly Archives: January 2014

Mistakes in trigonometry: easy, common, unexpected

We congratulate all Oxbridge candidates who have received their offers in January 2014!

In this post we discuss common mistakes in solving trigonometric equations.

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Easy mistake: basic values often forgotten

If you remember basic values of trigonometric functions such as \(\sin\frac{\pi}{6}\), \(\cos\frac{\pi}{4}\), \(\tan\frac{\pi}{3}\) without using a calculator, then at a real exam you will save time for more interesting problems.

A good exercise is to check the following basic values (for any integer n): \(\sin(n\pi)=0\), \(\sin(\frac{\pi}{2}+n\pi)=(-1)^n=\cos(n\pi)\), \(\tan(\frac{\pi}{4}+n\frac{\pi}{2})=(-1)^n=\cot(\frac{\pi}{4}+n\frac{\pi}{2})\). You may check these identities for n=0, n=1 (possibly for n=2, n=3) and then use periodicity.

A good style to finish your solution to a trigonometric equation is to substitute your answers back into the original equation. If something is wrong, please don’t write “something is wrong”, but try to correct your computations. In our distance courses we give more specific tips to locate a computational error in a long chain of equations.

Common mistake: not all solutions are found

We have seen hundreds of scripts that contain “\(\sin x=0\Rightarrow x=0\)“. This logical implication fails, because x=π also satisfies the equation \(\sin x=0\). Here is the full correct solution “\(\sin x=0, x=n\pi\) for any integer n”.

We wouldn’t write the arrow \(\Rightarrow\), because this logical implication doesn’t say that all found values are correct. For instance, the implication “\(\sin x=0 \Rightarrow x=n\frac{\pi}{2}\) for any integer n” is also logically correct. So this confusing notation will be a topic for one of our future posts.

Since trigonometric functions are periodic, a trigonometric equation usually has infinitely many solutions. To make problems simpler, the authors of MAT (entrance exams to Oxford and Imperial College London) often specify a short range of expected solutions, usually from 0 to 2π.

Unexpected mistake: Q1 in STEP III exam 2007

Here is the exact quote: “The four real numbers θ1, θ2, θ3 and θ4 lie in the range \(0\leq \theta_i< 2\pi\) and satisfy the equation \(p\cos(2\theta)+\cos(\theta-\alpha)+p=0\), where p and \(\alpha\) are independent of θ. Show that θ1234=nπ for some integer n.”

Let us consider the simple value of the parameter p=0. Then the given equation becomes \(\cos(\theta-\alpha)=0\) and has only two real solutions in the range \(0\leq\theta<2\pi\). For instance, if \(\alpha=\frac{\pi}{6}\), then the two solutions of \(\cos(\theta-\frac{\pi}{6})=0\) in the range \(0\leq\theta<2\pi\) are
\(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{5\pi}{3}\).

There is no way to assign these two values \(\frac{2\pi}{3}\) and \(\frac{5\pi}{3}\) to 4 variables whose sum θ1234 is a multiple of π. Indeed, if the value \(\frac{2\pi}{3}\) is taken k (of 4) times, then the value \(\frac{5\pi}{3}\) is taken 4-k times. Then \(k\frac{2\pi}{3}+(4-k)\frac{5\pi}{3}=n\pi\), hence \(2k+5(4-k)=3n\) and \(20=3(k+n)\), which is impossible for integer k,n.

A brute-force alternative is to consider all 5 cases for possible values of the 4 angles \(0\leq\theta_1\leq\theta_2\leq\theta_3\leq\theta_4<2\pi\) as follows.

(1) If \(\theta_1=\theta_2=\theta_3=\theta_4=\frac{2\pi}{3}\), then \(\theta_1+\theta_2+\theta_3+\theta_4=\frac{8\pi}{3}\).

(2) If \(\theta_1=\theta_2=\theta_3=\frac{2\pi}{3}\) and \(\theta_4=\frac{5\pi}{3}\), then \(\theta_1+\theta_2+\theta_3+\theta_4=\frac{11\pi}{3}\).

(3) If \(\theta_1=\theta_2=\frac{2\pi}{3}\) and \(\theta_3=\theta_4=\frac{5\pi}{3}\), then \(\theta_1+\theta_2+\theta_3+\theta_4=\frac{14\pi}{3}\).

(4) If \(\theta_1=\frac{2\pi}{3}\) and \(\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}\), then \(\theta_1+\theta_2+\theta_3+\theta_4=\frac{17\pi}{3}\).

(5) If \(\theta_1=\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}\), then \(\theta_1+\theta_2+\theta_3+\theta_4=\frac{20\pi}{3}\).

In all 5 cases (1)-(5) above the sum \(\theta_1+\theta_2+\theta_3+\theta_4\) is not a multiple of π, which disproves the statement from Q1 STEP III exam 2007. The STEP examiners’ solution to this problem is unfortunately too short and doesn’t consider exceptional cases when a denominator can be zero.

Rather surprisingly, neither the problem nor the solution were corrected in the official STEP resources for more than 6.5 years (checked on 24th January 2014). Fortunately, we have modified this problem and carefully considered all exceptional cases in our STEP III course.

  • Riddle 11: solve the equation \(\sin x=a\) for any real parameter a.
  • How to submit: to write your full answer, submit a comment.
  • Hint: there is a short formula for x depending on an integer n.
  • Warning: a full answer should cover all possible values of a.
  • Prize: free 1-year access to one of our interactive web tutorials.
  • Restriction: only the first correct public answer will be rewarded.
  • Update: K Wright has solved the problem, see attempts 1 and 2.

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