How to solve a difficult SSC CGL level problem in a few reasoned steps, Trigonometry 10 | SureSolv

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How to solve a difficult SSC CGL level problem in a few reasoned steps, Trigonometry 10

Identify key patterns and apply mathematical reasoning to solve in a few tens of seconds

How to solve a difficult trigonometry problem in a few reasoned steps 10

The chosen problem in this session is awkward and if you attempt to solve it through conventional deductions, it would be tedious and time consuming. You may even lose your way in confusion. Instead, if you are able to identify useful patterns and apply mathematical reasoning, solution would take just a few tens of seconds.

Before going ahead you may have look at our concept tutorials on Trigonometry,

Basic and Rich Trigonometry concepts and applications

Basic and Rich Trigonometry concepts part 2, Compound angle functions

Trigonometry concepts part 3, maxima and minima of trigonometric expressions.

Chosen problem.

The value of $\sin(B-C)\cos(A-D)$


$\hspace{22mm}+\sin(C-A)\cos(B-D)$ is,

  1. $\displaystyle\frac{3}{2}$
  2. $0$
  3. $-3$
  4. $1$

First solution: Conventional approach with deductive steps

We would expand each of the six compound angle $\sin$ functions and then simplify,




$=(\sin B\cos C-\cos B\sin C)(\cos A\cos D+\sin A\sin D)$

$\hspace{5mm}(\sin A\cos B-\cos A\sin B)(\cos C\cos D+\sin C\sin D)$

$\hspace{5mm}(\sin C\cos A-\cos C\sin A)(\cos B\cos D+\sin B\sin D)$

$=\cos A\sin B\cos C\cos D+\sin A\sin B\cos C\sin D$

$\hspace{5mm}-\cos A\cos B\sin C\cos D-\sin A\cos B\sin C\sin D$

$\hspace{5mm}+\sin A\cos B\cos C\cos D+\sin A\cos B\sin C\sin D$

$\hspace{5mm}-\cos A\sin B\cos C\cos D-\cos A\sin B\sin C\sin D$

$\hspace{5mm}+\cos A\cos B\sin C\cos D+\cos A\sin B\sin C\sin D$

$\hspace{5mm}-\sin A\cos B\cos C\cos D-\sin A\sin B\cos C\sin D$.

These are the 12 four factor terms, six positive and six negative. Carefully going through the terms we identify cancellation pairs as,

Term 1 and term 7; Term 2 and term 12; Term 3 and term 9; Term 4 and term 6; Term 5 and term 11; Term 8 and term 10. All terms accounted for and cancelling each other, the final result is 0.

Answer: Option c: $0$.

It is a purely deductive solution usually followed in schools. The difficulty in this problem arises at three levels—first, the two term expansions of six compound angle functions correctly needs concentration. Second—expansions of six two term factors resulting in 12 four factor terms again needs special attention; and the last—matching the 12 terms into cancelling pairs may be difficult unless the four angles in each of the 12 terms are ordered in the same manner. We have ordered it as A, B, C, D for all of the 12 terms.

Overall, this solution is impracticable in the context of an MCQ test. 

We will now present the elegant solution wholly based on key pattern identification and mathematical reasoning. For MCQ test, this is the only practicable solution for this problem.

Second solution: Problem analysis, Key pattern identification and Solution by mathematical reasoning

Each of the three product terms will expand to 4 unique terms because of the unique angle and function distribution. No term will be cancelled out within each product result and after expansion then, total number of terms will be 12.

The important pattern in these 12 terms identified is—each of the 12 terms consists of four unique factors of $\sin$ and $\cos$ using all the four angles.

Continuing our deductive reasoning it is obvious then,

All these 12 compound terms, each with four factors in $\sin$ and $\cos$, must cancel out against each other to generate the final result free of any variable.

Finally we reason that—if six of these twelve terms cancel out against the other six, the final result can only be 0, not any non-zero integer.

Answer: Option b: $0$.

Key concepts and techniques used:  Problem analysis -- Key pattern identification -- Deductive reasoning -- Mathematical reasoning -- Solving in mind.

Compare the shortcomings of the two solutions and identify the reasons behind the shortcomings.

Important to note

Elegant and efficient solutions are based on problem analysis, deductive reasoning and use of problem solving strategies and techniques. If you attempt to solve a problem using concepts in a few steps, gradually you will develop the skillset in reaching an elegant solution for most problems easily.

It is in a way, a habit of thinking in new ways.

Conversely, if you always follow the conventional path to the solution of a problem, you will never know how to think new and how to solve problems elegantly and efficiently.

Special characteristics of efficient trigonometry problem solving

  1. To achieve elegant and efficient solutions in Trigonometry consistently, role of basic and rich algebraic concepts and techniques is critically important.
  2. While solving the problem in two different ways, we have applied the problem solving skill improvement Many ways technique that tested our skill in solving a problem in multiple ways as well as gave us the opportunity to compare the multiple solutions with each other.

Even simple things need to be made simpler. Only then more difficult things can be made simple with ease.

Guided help on Trigonometry in Suresolv

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Reading and Practice Guide on Trigonometry in Suresolv for SSC CHSL, SSC CGL, SSC CGL Tier II Other Competitive exams.

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